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Introduction to Management Science
Introduction to
Management
Science
A Modeling and Case Studies Approach with Spreadsheets
Sixth Edition
Frederick S. Hillier
Stanford University
Mark S. Hillier
University of Washington
Cases developed by
Karl Schmedders
University of Zurich
Molly Stephens
Quinn, Emanuel, Urquhart & Sullivan, LLP
Final PDF to printer
INTRODUCTION TO MANAGEMENT SCIENCE
Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2019 by
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mheducation.com/highered
hiL91856_fm_ISE.indd
iv
12/11/17 02:27 PM
To the memory of
Christine Phillips Hillier
a beloved wife and daughter-in-law
Gerald J. Lieberman
an admired mentor and one of the true giants
of our field
About the Authors
Frederick S. Hillier is professor emeritus of operations research at Stanford University.
Dr. Hillier is especially known for his classic, award-winning text, Introduction to Operations
Research, co-authored with the late Gerald J. Lieberman, which has been translated into well
over a dozen languages and is currently in its 10th edition. The 6th edition won honorable mention for the 1995 Lanchester Prize (best English-language publication of any kind in the field),
and Dr. Hillier also was awarded the 2004 INFORMS Expository Writing Award for the 8th
edition. His other books include The Evaluation of Risky Interrelated Investments, Queueing
Tables and Graphs, Introduction to Stochastic Models in Operations Research, and Introduction to Mathematical Programming. He received his BS in industrial engineering and doctorate
specializing in operations research and management science from Stanford University. The winner of many awards in high school and college for writing, mathematics, debate, and music, he
ranked first in his undergraduate engineering class and was awarded three national f­ ellowships
(National Science Foundation, Tau Beta Pi, and Danforth) for graduate study. After receiving
his PhD degree, he joined the faculty of Stanford University, where he earned tenure at the age
of 28 and the rank of full professor at 32. Dr. Hillier’s research has extended into a variety of
areas, including integer programming, queueing theory and its application, statistical quality
control, and production and operations management. He also has won a major prize for research
in capital budgeting. Twice elected a national officer of professional societies, he has served in
many important professional and editorial capacities. For example, he served The Institute of
Management Sciences as vice president for meetings, chairman of the publications committee,
associate editor of Management Science, and co-general chairman of an international conference in Japan. He also is a Fellow of the Institute for Operations Research and the Management Sciences (INFORMS). He served for 20 years (until 2013) as the founding series editor
for a prominent book series, the International Series in Operations Research and ­Management
­Science, for Springer Science + Business Media. He has had visiting appointments at Cornell
University, the Graduate School of Industrial Administration of Carnegie-Mellon University,
the Technical University of Denmark, the University of Canterbury (New Zealand), and the
Judge Institute of Management Studies at the University of Cambridge (England).
vi
Mark S. Hillier, son of Fred Hillier, is associate professor of quantitative methods at the Michael G.
Foster School of Business at the University of Washington. Dr. Hillier received his BS in engineering (plus a concentration in computer science) from Swarthmore College. He then received
his MS with distinction in operations research and PhD in industrial engineering and engineering management from Stanford University. As an undergraduate, he won the McCabe Award
for ranking first in his engineering class, won election to Phi Beta Kappa based on his work in
mathematics, set school records on the men’s swim team, and was awarded two national fellowships (National Science Foundation and Tau Beta Pi) for graduate study. During that time, he also
developed a comprehensive software tutorial package, OR Courseware, for the Hillier–Lieberman
textbook, Introduction to Operations Research. As a graduate student, he taught a PhD-level seminar in operations management at Stanford and won a national prize for work based on his PhD dissertation. At the University of Washington, he currently teaches courses in management science
and spreadsheet modeling. He has won over twenty MBA teaching awards for the core course in
management science and his elective course in spreadsheet modeling, as well as a universitywide
teaching award for his work in teaching undergraduate classes in operations management. He was
chosen by MBA students in both 2007 and 2013 as the winner of the prestigious PACCAR award
for Teacher of the Year (reputed to provide the largest monetary award for MBA teaching in the
nation). He won the Ron Crocket Award for Innovation in Education in 2014. He also has been
awarded an appointment to the Evert McCabe Endowed Faculty Fellowship. His research interests
include issues in component commonality, inventory, manufacturing, and the design of production systems. A paper by Dr. Hillier on component commonality won an award for best paper of
2000–2001 in IIE Transactions. He also has been serving as principal investigator on a grant from
the Bill and Melinda Gates Foundation to lead student research projects that apply spreadsheet
modeling to various issues in global health being studied by the foundation.
About the Case Writers
Karl Schmedders is professor of quantitative business administration at the University of
Zurich in Switzerland and a visiting professor of executive education at the Kellogg School
of Management of Northwestern University. His research interests include management science, business analytics, and computational economics and finance. He received his PhD
in operations research from Stanford University, where he taught both undergraduate and
graduate classes in management science, including a case studies course. He received several
teaching awards at Stanford, including the universitywide Walter J. Gores Teaching Award.
After a post-doc at the Hoover Institution, a think tank on the Stanford campus, he became
assistant professor of managerial economics and decision sciences at the Kellogg School. He
was promoted to associate professor in 2001 and received tenure in 2005. In 2008 he joined
the University of Zurich, where he currently teaches courses in management science, business
analytics, and computational economics and finance. He has published research articles in
international academic journals such as Management Science, Operations Research, Econometrica, The Review of Economic Studies, and The Journal of Finance, among others. He is
the area editor of the field “Computational Economics” for the INFORMS journal Operations
Research. At Kellogg he received several teaching awards, including the L. G. Lavengood
Professor of the Year Award. Most recently he won the best professor award of the Kellogg
School’s European EMBA program (2011, 2012, 2013, 2014, 2015) and its EMBA program
in Hong Kong (2017).
Molly Stephens is a partner in the Los Angeles office of Quinn, Emanuel, Urquhart &
­Sullivan, LLP. She graduated from Stanford with a BS in industrial engineering and an MS in
operations research. Ms. Stephens taught public speaking in Stanford’s School of E
­ ngineering
and served as a teaching assistant for a case studies course in management science. As a teaching assistant, she analyzed management science problems encountered in the real world and
transformed these into classroom case studies. Her research was rewarded when she won an
undergraduate research grant from Stanford to continue her work and was invited to speak at
INFORMS to present her conclusions regarding successful classroom case studies. ­Following
graduation, Ms. Stephens worked at Andersen Consulting as a systems integrator, experiencing real cases from the inside, before resuming her graduate studies to earn a JD degree with
honors from the University of Texas School of Law at Austin. She is a partner in the largest
law firm in the United States devoted solely to business litigation, where her practice focuses
on complex financial and securities litigation. She also is ranked as a leading securities litigator by Chambers USA (2013 and 2014), which acknowledged “praise for her powerful and
impressive securities litigation practice” and noted that she is “phenomenally bright, a critical
thinker and great listener.”
vii
Preface
We have long been concerned that traditional management science textbooks have not taken
the best approach in introducing business students to this exciting field. Our goal when initially developing this book during the late 1990s was to break out of the old mold and present
new and innovative ways of teaching management science more effectively. We have been
gratified by the favorable response to our efforts. Many reviewers and other users of the first
five editions of the book have expressed appreciation for its various distinctive features, as
well as for its clear presentation at just the right level for their business students.
Our goal for this sixth edition has been to build on the strengths of the first five editions.
Co-author Mark Hillier has won over twenty schoolwide teaching awards for his spreadsheet
modeling and management science courses at the University of Washington while using the
first five editions, and this experience has led to many improvements in the current edition.
We also incorporated many user comments and suggestions. Throughout this process, we
took painstaking care to enhance the quality of the preceding edition while maintaining the
distinctive orientation of the book.
This distinctive orientation is one that closely follows the recommendations in the 1996
report of the operating subcommittee of the INFORMS Business School Education Task
Force, including the following extract.
There is clear evidence that there must be a major change in the character of the (introductory
management science) course in this environment. There is little patience with courses centered
on algorithms. Instead, the demand is for courses that focus on business situations, include
prominent non-mathematical issues, use spreadsheets, and involve model formulation and
assessment more than model structuring. Such a course requires new teaching materials.
This book is designed to provide the teaching materials for such a course.
In line with the recommendations of this task force, which continue to be widely accepted
today, we believe that a modern introductory management science textbook should have three
key elements. As summarized in the subtitle of this book, these elements are a modeling and
case studies approach with spreadsheets.
SPREADSHEETS
The modern approach to the teaching of management science clearly is to use spreadsheets
as a primary medium of instruction. Both business students and managers now live with
spreadsheets, so they provide a comfortable and enjoyable learning environment. Modern
spreadsheet software, including Microsoft Excel used in this book, now can be used to do
real management science. For student-scale models (which include many practical real-world
models), spreadsheets are a much better way of implementing management science models
than traditional algebraic solvers. This means that the algebraic curtain that continues to be
prevalent in traditional management science courses and textbooks now can be lifted.
However, with the current enthusiasm for spreadsheets, there is a danger of going overboard. Spreadsheets are not the only useful tool for performing management science analyses. Occasional modest use of algebraic and graphical analyses still have their place and
we would be doing a disservice to the students by not developing their skills in these areas
when appropriate. Furthermore, the book should not be mainly a spreadsheet cookbook that
focuses largely on spreadsheet mechanics. Spreadsheets are a means to an end, not an end in
themselves.
A MODELING APPROACH
This brings us to the second key feature of the book, a modeling approach. Model formulation
lies at the heart of management science methodology. Therefore, we heavily emphasize the art
of model formulation, the role of a model, and the analysis of model results. We primarily (but
not exclusively) use a spreadsheet format rather than algebra for formulating and presenting
a model.
viii
Preface ix
Some instructors have many years of experience in teaching modeling in terms of formulating algebraic models (or what the INFORMS Task Force called “model structuring”).
Some of these instructors feel that students should do their modeling in this way and then
transfer the model to a spreadsheet simply to use the Excel Solver to solve the model. We disagree with this approach. Our experience (and the experience reported by many others) is that
most business students find it more natural and comfortable to do their modeling directly in
a spreadsheet. Furthermore, by using the best spreadsheet modeling techniques (as presented
in this edition), formulating a spreadsheet model tends to be considerably more efficient
and transparent than formulating an algebraic model. Another benefit is that the spreadsheet
model includes all the relationships that can be expressed in an algebraic form and we often
will summarize the model in this format as well.
Another break from past tradition in this book (and several contemporary textbooks) is
to virtually ignore the algorithms that are used to solve the models. We feel that there is no
good reason why typical business students should learn the details of algorithms executed
by computers. Within the time constraints of a one-term management science course, there
are far more important lessons to be learned. Therefore, the focus in this book is on what we
believe are these far more important lessons. High on this list is the art of modeling managerial problems on a spreadsheet.
We believe that training business students in spreadsheet modeling will provide them with
two key benefits when they later become managers. First, this will give them a powerful tool
for analyzing small managerial problems without requiring outside help. Second, this will
enable them to recognize when a management science team could be very helpful for analyzing more complicated managerial problems.
Formulating a spreadsheet model of a real problem typically involves much more than
designing the spreadsheet and entering the data. Therefore, we work through the process
step by step: understand the unstructured problem, verbally develop some structure for the
problem, gather the data, express the relationships in quantitative terms, and then lay out the
spreadsheet model. The structured approach highlights the typical components of the model
(the data, the decisions to be made, the constraints, and the measure of performance) and the
different types of spreadsheet cells used for each. Consequently, the emphasis is on the modeling rather than spreadsheet mechanics.
A CASE STUDIES APPROACH
However, all this still would be quite sterile if we simply presented a long series of brief
examples with their spreadsheet formulations. This leads to the third key feature of this
book—a case studies approach. In addition to examples, nearly every chapter includes one or
two case studies patterned after actual applications to convey the whole process of applying
management science. In a few instances, the entire chapter revolves around a case study. By
drawing the student into the story, we have designed each case study to bring that chapter’s
technique to life in a context that vividly illustrates the relevance of the technique for aiding
managerial decision making. This storytelling, case-centered approach should make the material more enjoyable and stimulating while also conveying the practical considerations that are
key factors in applying management science.
We have been pleased to have several reviewers of the first five editions express particular appreciation for our case study approach. Even though this storytelling approach
has received little use in other management science textbooks, we feel that it is a real key
to preparing students for the practical application of management science in all its aspects.
Some of the reviewers have highlighted the effectiveness of the dialogue/scenario enactment approach used in some of the case studies. Although unconventional, this approach
provides a way of demonstrating the process of managerial decision making with the help
of management science. It also enables previewing some key concepts in the language of
management.
Every chapter also contains full-fledged cases following the problems at the end of the
chapter. These cases usually continue to employ a stimulating storytelling approach, so they
can be assigned as interesting and challenging projects. Most of these cases were developed
x Preface
jointly by two talented case writers, Karl Schmedders (a faculty member at the University of
Zurich in Switzerland) and Molly Stephens (formerly a management science consultant with
Andersen Consulting). The authors also have added some cases, including several shorter
ones. In addition, the University of Western Ontario Ivey School of Business (the secondlargest producer of teaching cases in the world) has specially selected cases from their case
collection that match the chapters in this textbook. These cases are available on the Ivey website, cases.ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book.
This website address is provided at the end of each chapter as well.
We are, of course, not the first to incorporate any of these key features into a management
science textbook. However, we believe that the book currently is unique in the way that it
fully incorporates all three key features together.
OTHER SPECIAL FEATURES
We also should mention some additional special features of the book that are continued from
the fifth edition.
• Diverse examples, problems, and cases convey the pervasive relevance of management
science.
• A strong managerial perspective.
• Learning objectives at the beginning of each chapter.
• Numerous margin notes that clarify and highlight key points.
• Excel tips interspersed among the margin notes.
• Review questions at the end of each section.
• A glossary at the end of each chapter.
• Partial answers to selected problems in the back of the book.
• Extensive supplementary text materials (nine supplements to book chapters and seven additional chapters) are available in the Instructor Resources of McGraw-Hill’s online homework management platform, Connect, (described later) and on the website, www.mhhe
.com/Hillier6e.
A NEW SOFTWARE PACKAGE
This edition continues to integrate Excel and its Solver (a product originally created by Frontline Systems) throughout the book. However, we are excited to also add to this edition an
impressive more recent product of Frontline Systems called Analytic Solver ® for ­Education
(hereafter referred to as Analytic Solver). Analytic Solver can be used with Microsoft Excel
for Windows (as an add-in), or ‘in the cloud” at AnalyticSolver.com using any device (PC,
Mac, tablet) with a web browser. It offers comprehensive features for prescriptive analytics
(optimization, simulation, decision analysis) and predictive analytics (forecasting, data mining, text mining). Its optimization features are upward compatible from the standard Solver in
Excel. Analytic Solver includes:
• A more interactive user interface, with the model parameters always visible alongside the
main spreadsheet, rather than only in the Solver dialog box.
• Parameter analysis reports that provide an easy way to see the effect of varying data in a
model in a systematic way.
• A model analysis tool that reveals the characteristics of a model (e.g., whether it is linear
or nonlinear, smooth or nonsmooth).
• Tools to build and solve decision trees within a spreadsheet.
• A full range of time series forecasting and data mining models.
• The ability to build and run sophisticated Monte Carlo simulation models.
• An interactive simulation mode that allows simulation results to be shown instantly whenever a change is made to a simulation model.
• The Solver in Analytic Solver can be used in combination with computer simulation to
perform simulation optimization.
Preface xi
• If interested in having students get individual licenses for class use, instructors should
send an email to support@solver.com to get their course code and receive student pricing and access information as well as their own access information. Note that this software
is no longer free with the purchase of this text, but low-cost student licenses are available.
A CONTINUING FOCUS ON EXCEL AND ITS SOLVER
As with all the preceding editions, this edition continues to focus on spreadsheet modeling in
an Excel format. Although it lacks some of the functionalities of Analytic Solver, the Excel
Solver continues to provide a completely satisfactory way of solving most of the spreadsheet
models encountered in this book. This edition continues to feature this use of the Excel Solver
whenever either it or the Analytic Solver could be used.
Many instructors prefer this focus because it avoids introducing other complications that
might confuse their students. We agree.
However, the key advantage of introducing Analytic Solver in this edition is that it provides an all-in-one complement to the Excel Solver. There are some important topics in the
book (including decision analysis and computer simulation) where the Excel Solver lacks the
functionalities needed to deal fully with these kinds of problems. Multiple Excel add-ins—
Solver Table, TreePlan, SensIt, RiskSim, Crystal Ball, and OptQuest (a module of Crystal
Ball)—were introduced in previous editions to provide the needed functionalities. Analytic
Solver alone now replaces all of these add-ins.
To further enhance a continuing focus on Excel and its Solver, www.mhhe.com/Hillier6e
includes all the Excel files that provide the live spreadsheets for all the various examples and
case studies throughout the book. In addition to further investigating the examples and case
studies, these spreadsheets can be used by either the student or instructor as templates to formulate and solve similar problems. This website also includes dozens of Excel templates for
solving various models in the book as well as a Queueing Simulator for performing computer
simulations of queueing systems (used in Chapter 12).
NEW FEATURES IN THIS EDITION
We have made some important enhancements to the current edition.
• A New Section Describes the Relationship Between Analytics and Management Science.
Recent years have seen an exciting analytics revolution as the business world has come to
recognize the key role that analytics can play in managerial decision making. A new ­Section
1.3 fully describes the close relationship between analytics and management science.
• A New Section on the Role of Robust Optimization in What-If Analysis. The goal of
robust optimization is to find a solution for a model that is virtually guaranteed to remain
feasible and near optimal for all plausible combinations of the actual values for the parameters of the model. A new Section 5.7 describes this key tool for performing what-if analysis for linear programming models.
• A New Section on Chance Constraints. Constraints in a model usually are fully required
to be satisfied, but occasionally constraints arise that actually have a little flexibility.
Chance constraints provide a way of dealing with such constraints that actually can be
violated a little bit without serious complications. A new Section 5.8 describes the role of
chance constraints and how to implement them.
• A Thorough Updating Throughout the Book. Given that the writing of the first edition occurred approximately 20 years ago, it is inevitable that some of that writing now is
somewhat outdated. Even though the descriptions of management science techniques may
still be accurate, the numbers and other details describing their application in certain problems, cases, and examples may now seem quite obsolete. Wage standards have changed.
Prices have changed. Technologies have changed. Dates have changed. Although we did
some updating with each new edition, we made a special effort this time to thoroughly
update numbers and other details as needed to reflect conditions in 2017.
• Additional Links to Articles that Describe Dramatic Real Applications. The fifth edition
included 28 application vignettes that described in a few paragraphs how an actual application of management science had a powerful effect on a company or organization by using
xii Preface
techniques like those being studied in that portion of the book. The current edition adds five
more vignettes based on recent applications (while deleting seven outdated ones) and also
updates the information in nine of the other vignettes. We continue the practice of adding a
link to the journal articles that fully describe these applications (except that the vignette for
Chapter 1 doesn’t require a link), through a special arrangement with the Institute for Operations Research and the Management Sciences (INFORMS®). Thus, the instructor now can
motivate his or her lectures by having the students delve into real applications that dramatically demonstrate the relevance of the material being covered in the lectures. The end-ofchapter problems also include an assignment after reading each of these articles.
We continue to be excited about this partnership with INFORMS, our field’s preeminent professional society, to provide a link to each of these articles describing spectacular applications of management science. INFORMS is a learned professional society for
students, academics, and practitioners in analytics, operations research, and management
science. Information about INFORMS journals, meetings, job bank, scholarships, awards,
and teaching materials is available at www.informs.org.
• A Word-by-Word Review to Further Increase Clarity in Each Chapter. A hallmark
of each edition has been a particularly heavy use of certain techniques to maximize the
clarity of the material: use cases to bring the material to life, divide sections into smaller
subsections, use short paragraphs, use bullet points, set off special conclusions, use i­talics
or boldface to highlight key points, add margin notes, never assume too much about
­understanding preceding material, etc. However, we have doubled down with this approach
in the current edition by using a word-by-word review of each chapter to further increase
clarity while also taking into special account the input provided by reviewers and others.
McGRAW-HILL CONNECT(R) LEARN WITHOUT LIMITS!
Connect is a teaching and learning platform that is proven to deliver better results for students and
instructors. Connect empowers students by continually adapting to deliver precisely what they
need, when they need it, and how they need it, so your class time is more engaging and effective.
New to the 6th edition, Connect includes multiple-choice questions for each chapter to be
used as practice or homework for students, SmartBook(R), instructor resources (including the
test bank), and student resources. For access, visit connect.mheducation.com or contact your
McGraw-Hill sales representative.
SMARTBOOK(R)
Proven to help students improve grades and study more efficiently, SmartBook contains the
same content within the print book, but actively tailors that content to the needs of the individual. SmartBook’s adaptive technology provides precise, personalized instruction on what
the student should do next, guiding the student to master and remember key concepts, targeting gaps in knowledge and offering customized feedback, and driving the student toward
comprehension and retention of the subject matter. Available on desktops and tablets, SmartBook puts learning at the student’s fingertips—anywhere, anytime.
INSTRUCTOR RESOURCES
The Instructor Resource Library within Connect is password-protected and a convenient place
for instructors to access course supplements that include nine supplements to chapters in the
print book and seven supplementary chapters. Resources for professors include the complete
solutions to all problems and cases, PowerPoint slides which include both lecture materials for
nearly every chapter and nearly all the figures (including all the spreadsheets) in the book, and
an expanded Test Bank. The test bank contains almost 1,000 multiple-choice and true-false
questions, all tagged according to learning objective, topic, level of difficulty, Bloom’s taxonomy and AACSB category for filtering and reporting, and in delivered in the following ways:
- As a Connect assignment for online testing and automatic grading; can be used for actual
exams or assigned as quizzes or practice;
- In TestGen, a desktop test generator and editing application for instructors to provide
printed tests that can incorporate both McGraw-Hill’s and instructors’ questions;
- As Word files, with both question-only and answer files.
Final PDF to printer
Preface xiii
STUDENT RESOURCES
As described above, SmartBook provides a powerful tool to students for personalized instruction.
For the additional convenience of students, we also are providing the website, www.mhhe.com/
Hillier6e, to provide the full range of resources of interest to students. In addition to providing
access to supplementary text material (both supplements to book chapters and additional chapters), this website provides solutions to the “solved problems” (additional examples) that are
included at the end of each chapter. For each spreadsheet example in the book, a live spreadsheet
that shows the formulation and solution for the example also is provided in the website for easy
reference and for use as a template. (At the end of each chapter, the page entitled “Learning
Aids for This Chapter” lists the Excel files and other resources that are relevant for that chapter.)
Information about accessing the book’s software is provided. In addition, the website includes
a tutorial with sample test questions (different from those in the instructor’s test bank) for selftesting quizzes on the various chapters. It also provides access to the INFORMS articles cited in
the application vignettes as well as updates about the book, including errata.
AN INVITATION
We invite your comments, suggestions, and errata. You can contact either one of us at the
e-mail addresses given below. While giving these addresses, let us also assure instructors that
we will continue our policy of not providing solutions to problems and cases in the book to
anyone (including your students) who contacts us. We hope that you enjoy the book.
Frederick S. Hillier
Stanford University (fhillier@stanford.edu)
Mark S. Hillier
University of Washington (mhillier@uw.edu)
June 2017
Acknowledgments
This new edition has benefited greatly from the sage advice of many individuals. To begin, we would like to express our
deep appreciation to the following individuals who provided formal reviews of the fifth edition:
Michael Cervertti,
University of Memphis
Jose H. Dula,
Virginia Commonwealth University
John H. Newman,
Coppin State University
Danxia Chen,
Dallas Baptist University
Kenneth Donald Lawrence,
New Jersey Institute of Technology
David Snowden,
Thomas More College
Moula Cherikh,
Winston-Salem State University
Nicoleta Maghear,
Hampton University
John Wang,
Montclair State University
Kelwyn A. D’Souza,
Hampton University
William P. Millhiser,
Baruch College, City University of New York
Norman Douglas Ward,
University of Mount Olive
We also are grateful for the valuable input provided by many of our students as well as various other students and
instructors who contacted us via e-mail.
This book has continued to be a team effort involving far more than the two coauthors. As a third coauthor for the first
edition, the late Gerald J. Lieberman provided important initial impetus for this project. We also are indebted to our case writers, Karl Schmedders and Molly Stephens, for their invaluable contributions. Ann Hillier again devoted numerous hours to
sitting with a Macintosh, doing word processing and constructing figures and tables. They all were vital members of the team.
McGraw-Hill/Irwin’s editorial and production staff provided the other key members of the team, including Noelle
Bathurst, Portfolio Manager; Allison McCabe, and Tobi Philips, Product Developers, Harper Christopher, Marketing
Manager, and Daryl Horrocks, Program Manager. This book is a much better product because of their guidance and hard
work. It has been a real pleasure working with such a thoroughly professional staff.
hiL18920_fm_i-xviii.indd
xiii
12/20/17 09:44 AM
Brief Contents
1 Introduction 1
10 Forecasting 397
2 Linear Programming: Basic Concepts 24
11 Queueing Models 446
3 Linear Programming: Formulation and
Applications 64
12 Computer Simulation: Basic Concepts 499
4 The Art of Modeling with
Spreadsheets 124
5 What-If Analysis for Linear
Programming 151
6 Network Optimization Problems 203
7 Using Binary Integer Programming to Deal
with Yes-or-No Decisions 243
8 Nonlinear Programming 279
13 Computer Simulation with Analytic
Solver 536
APPENDIXES
A
Tips for Using Microsoft Excel for
Modeling 606
B
Partial Answers to Selected Problems 612
INDEX 616
9 Decision Analysis 334
CHAPTERS available at www.mhhe.com/Hillier6e
14
Solution Concepts for Linear Programming
15
Transportation and Assignment Problems
16
PERT/CPM Models for Project
Management
17
Goal Programming
xiv
18
Inventory Management with Known
Demand
19
Inventory Management with Uncertain
Demand
20
Computer Simulation with Crystal Ball
Contents
Chapter One
Introduction 1
The Nature of Management Science 2
An Illustration of the Management Science
Approach: Break-Even Analysis 6
1.3
The Relationship Between Analytics and
Management Science 12
1.4
The Impact of Management Science 14
1.5
Some Special Features of this Book 18
1.6
Summary 19
Glossary 20
Learning Aids for This Chapter 20
Solved Problem 21
Problems 21
Case 1-1 Keeping Time 23
1.1
1.2
Chapter Two
Linear Programming: Basic Concepts 24
A Case Study: The Wyndor Glass Co.
Product-Mix Problem 25
2.2
Formulating the Wyndor Problem on a
Spreadsheet 27
2.3
The Mathematical Model in the Spreadsheet 33
2.4
The Graphical Method for Solving Two-Variable
Problems 35
2.5
Using Excel’s Solver to Solve Linear
Programming Problems 39
2.6
Analytic Solver 43
2.7
A Minimization Example—The Profit &
Gambit Co. Advertising-Mix Problem 47
2.8
Linear Programming from a Broader Perspective 52
2.9
Summary 54
Glossary 54
Learning Aids for This Chapter 55
Solved Problems 55
Problems 55
Case 2-1 Auto Assembly 60
Case 2-2 Cutting Cafeteria Costs 61
Case 2-3 Staffing a Call Center 62
2.1
Supplement to Chapter 2: More about the Graphical
Method for Linear Programming (This supplement is available at www.mhhe.com/Hillier6e).
Chapter Three
Linear Programming: Formulation and
Applications 64
3.1
3.2
3.3
3.4
3.5
3.6
A Case Study: The Super Grain Corp.
Advertising-Mix Problem 65
Resource-Allocation Problems 71
Cost–Benefit–Trade-Off Problems 81
Mixed Problems 87
Transportation Problems 95
Assignment Problems 99
3.7
Model Formulation from a Broader Perspective 102
3.8
Summary 104
Glossary 104
Learning Aids for This Chapter 104
Solved Problems 105
Problems 106
Case 3-1 Shipping Wood to Market 114
Case 3-2 Capacity Concerns 115
Case 3-3 Fabrics and Fall Fashions 117
Case 3-4 New Frontiers 118
Case 3-5 Assigning Students to Schools 119
Case 3-6 Reclaiming Solid Wastes 120
Case 3-7 Project Pickings 121
Chapter Four
The Art of Modeling with Spreadsheets 124
A Case Study: The Everglade Golden Years Company Cash Flow Problem 125
4.2
Overview of the Process of Modeling with
Spreadsheets 126
4.3
Some Guidelines for Building “Good”
Spreadsheet Models 136
4.4
Debugging a Spreadsheet Model 142
4.5
Summary 146
Glossary 146
Learning Aids for This Chapter 146
Solved Problems 146
Problems 147
Case 4-1 Prudent Provisions for Pensions 150
4.1
Chapter Five
What-If Analysis for Linear Programming 151
The Importance of What-If Analysis to Managers 152
Continuing the Wyndor Case Study 154
The Effect of Changes in One Objective Function
Coefficient 156
5.4
The Effect of Simultaneous Changes in Objective
­Function Coefficients 162
5.5
The Effect of Single Changes in a Constraint 169
5.6
The Effect of Simultaneous Changes in the
Constraints 175
5.7
Robust Optimization 179
5.8
Chance Constraints with Analytic Solver 182
5.9
Summary 186
Glossary 186
Learning Aids for This Chapter 187
Solved Problem 187
Problems 188
Case 5-1 Selling Soap 197
Case 5-2 Controlling Air Pollution 198
Case 5-3 Farm Management 200
Case 5-4 Assigning Students to Schools (Revisited) 202
5.1
5.2
5.3
Supplement to Chapter 5: Reduced Costs (This
­supplement is available at www.mhhe.com/Hillier6e).
xv
xvi Contents
Chapter Six
Network Optimization Problems 203
Minimum-Cost Flow Problems 204
A Case Study: The BMZ Co. Maximum
Flow Problem 212
6.3
Maximum Flow Problems 215
6.4
Shortest Path Problems 219
6.5
Summary 229
Glossary 229
Learning Aids for This Chapter 230
Solved Problems 230
Problems 231
Case 6-1 Aiding Allies 235
Case 6-2 Money in Motion 238
Case 6-3 Airline Scheduling 240
Case 6-4 Broadcasting the Olympic Games 241
6.1
6.2
Supplement to Chapter 6: Minimum Spanning-Tree Problems (This supplement is available at www.mhhe.com
/Hillier6e).
Chapter Seven
Using Binary Integer Programming to Deal with
Yes-or-No Decisions 243
A Case Study: The California Manufacturing Co.
Problem 244
7.2
Using BIP for Project Selection: The Tazer Corp.
Problem 251
7.3
Using BIP for The Selection of Sites for Emergency
Services Facilities: The Caliente City Problem 253
7.4
Using BIP for Crew Scheduling: The Southwestern
Airways Problem 257
7.5
Using Mixed BIP to Deal with Setup Costs for Initiating Production: The Revised Wyndor Problem 261
7.6
Summary 266
Glossary 266
Learning Aids for This Chapter 266
Solved Problems 266
Problems 268
Case 7-1 Assigning Art 273
Case 7-2 Stocking Sets 275
Case 7-3 Assigning Students to Schools (Revisited) 278
Case 7-4 Broadcasting the Olympic Games
(Revisited) 278
7.1
Supplement 1 to Chapter 7: Advanced Formulation Techniques for Binary Integer Programming
Supplement 2 to Chapter 7: Some Perspectives on Solving
Binary Integer Programming Problems (These supplements
are available at www.mhhe.com/Hillier6e.)
Chapter Eight
Nonlinear Programming 279
8.1
8.2
8.3
8.4
8.5
The Challenges of Nonlinear Programming 281
Nonlinear Programming with Decreasing ­Marginal
Returns 289
Separable Programming 299
Difficult Nonlinear Programming Problems 309
Evolutionary Solver and Genetic Algorithms 310
Using Analytic Solver to Analyze a Model
and Choose a Solving Method 318
8.7
Summary 322
Glossary 323
Learning Aids for This Chapter 323
Solved Problem 324
Problems 324
Case 8-1 Continuation of the Super Grain Case
Study 329
Case 8-2 Savvy Stock Selection 330
Case 8-3 International Investments 331
8.6
Chapter Nine
Decision Analysis 334
A Case Study: The Goferbroke Company
Problem 335
9.2
Decision Criteria 337
9.3
Decision Trees 342
9.4
Sensitivity Analysis with Decision Trees 346
9.5
Checking Whether to Obtain More Information 350
9.6
Using New Information to Update the
Probabilities 353
9.7
Using a Decision Tree to Analyze the Problem with
a Sequence of Decisions 357
9.8
Performing Sensitivity Analysis on the Problem
with a Sequence of Decisions 364
9.9
Using Utilities to Better Reflect the Values of
Payoffs 367
9.10 The Practical Application of Decision Analysis 378
9.11 Summary 379
Glossary 379
Learning Aids for This Chapter 380
Solved Problems 381
Problems 381
Case 9-1 Who Wants to Be a Millionaire? 391
Case 9-2 University Toys and the Business Professor
Action Figures 392
Case 9-3 Brainy Business 393
Case 9-4 Smart Steering Support 394
9.1
Supplement 1 to Chapter 9: Decision Criteria
Supplement 2 to Chapter 9: Using TreePlan Software for
Decision Trees (These supplements are available at www
.mhhe.com/Hillier6e.)
Chapter Ten
Forecasting 397
An Overview of Forecasting Techniques 398
A Case Study: The Computer Club Warehouse
(CCW) Problem 400
10.3 Applying Time-Series Forecasting Methods to the
Case Study 404
10.4 The Time-Series Forecasting Methods in
Perspective 423
10.5 Causal Forecasting with Linear Regression 426
10.6 Judgmental Forecasting Methods 431
10.7 Summary 433
Glossary 433
Summary of Key Formulas 434
10.1
10.2
Contents xvii
Learning Aids for This Chapter 435
Solved Problem 435
Problems 435
Case 10-1 Finagling the Forecasts 442
Chapter Eleven
Queueing Models 446
Elements of a Queueing Model 447
Some Examples of Queueing Systems 453
Measures of Performance for Queueing
Systems 455
11.4 A Case Study: The Dupit Corp. Problem 458
11.5 Some Single-Server Queueing Models 461
11.6 Some Multiple-Server Queueing Models 469
11.7 Priority Queueing Models 474
11.8 Some Insights about Designing Queueing
Systems 480
11.9 Economic Analysis of the Number of Servers
to Provide 485
11.10 Summary 488
Glossary 489
Key Symbols 490
Learning Aids for This Chapter 490
Solved Problem 490
Problems 491
Case 11-1 Queueing Quandary 497
Case 11-2 Reducing In-Process Inventory 498
11.1
11.2
11.3
Supplement to Chapter 11: Additional Queueing Models
(This supplement is available at www.mhhe.com/Hillier6e.)
Chapter Twelve
Computer Simulation: Basic Concepts 499
The Essence of Computer Simulation 500
A Case Study: Herr Cutter’s Barber Shop
(Revisited) 512
12.3 Analysis of the Case Study 519
12.4 Outline of a Major Computer Simulation Study 526
12.5 Summary 529
Glossary 529
12.1
12.2
Learning Aids for This Chapter 530
Solved Problem 530
Problems 530
Case 12-1 Planning Planers 534
Case 12-2 Reducing In-Process Inventory (Revisited) 535
Supplement to Chapter 12: The Inverse Transformation
Method for Generating Random Observations (This supplement is available at www.mhhe.com/Hillier6e.)
Chapter Thirteen
Computer Simulation with Analytic Solver 536
A Case Study: Freddie the Newsboy’s Problem 537
Bidding for a Construction Project: A Prelude
to the Reliable Construction Co. Case Study 547
13.3 Project Management: Revisiting the Reliable Construction Co. Case Study 551
13.4 Financial Risk Analysis: Revisiting the
Think-Big Development Co. Problem 557
13.5 Revenue Management in the Travel Industry 562
13.6 Choosing the Right Distribution 568
13.7 Decision Making with Parameter Analysis Reports
and Trend Charts 579
13.8 Optimizing with Computer Simulation Using
the Solver in Analytic Solver 587
13.9 Summary 595
Glossary 596
Learning Aids for This Chapter 596
Solved Problem 596
Problems 597
Case 13-1 Action Adventures 602
Case 13-2 Pricing under Pressure 603
Case 13-3 Financial Planning for Retirement 604
13.1
13.2
Appendix A: Tips for Using Microsoft Excel for
Modeling 606
Appendix B: Partial Answers to Selected
Problems 612
Index 616
CHAPTERS available at www.mhhe.com/Hillier6e
Chapter Fourteen
Solution Concepts for Linear Programming
Some Key Facts about Optimal Solutions
The Role of Corner Points in Searching for an Optimal Solution
14.3 Solution Concepts for the Simplex Method
14.4 The Simplex Method with Two Decision Variables
14.5 The Simplex Method with Three Decision Variables
14.6 The Role of Supplementary Variables
14.7 Some Algebraic Details for the Simplex Method
14.8 Computer Implementation of the Simplex Method
14.9 The Interior-Point Approach to Solving Linear Programming Problems
14.10 Summary
14.1
14.2
Glossary
Learning Aids for This Chapter
Problems
Chapter Fifteen
Transportation and Assignment Problems
15.1
15.2
15.3
15.4
15.5
A Case Study: The P & T Company Distribution
Problem
Characteristics of Transportation Problems
Modeling Variants of Transportation Problems
Some Other Applications of Variants of Transportation Problems
A Case Study: The Texago Corp. Site Selection
Problem
xviii Contents
15.6 Characteristics of Assignment Problems
15.7 Modeling Variants of Assignment Problems
15.8 Summary
Glossary
Learning Aids for This Chapter
Problems
Case 15-1 Continuation of the Texago Case Study
18.7 The EOQ Model with Gradual Replenishment
18.8 Summary
Glossary
Learning Aids for This Chapter
Problems
Case 18-1 Brushing Up on Inventory Control
Chapter Sixteen
Inventory Management with Uncertain Demand
PERT/CPM Models for Project Management
A Case Study: The Reliable Construction Co. Project
Using a Network to Visually Display a Project
Scheduling a Project with PERT/CPM
Dealing with Uncertain Activity Durations
Considering Time–Cost Trade-Offs
Scheduling and Controlling Project Costs
An Evaluation of PERT/CPM from a Managerial
Perspective
16.8 Summary
Glossary
Learning Aids for This Chapter
Problems
Case 16-1 Steps to Success
Case 16-2 “School’s Out Forever . . .”
16.1
16.2
16.3
16.4
16.5
16.6
16.7
Chapter Seventeen
Goal Programming
A Case Study: The Dewright Co. Goal-Programming Problem
17.2 Weighted Goal Programming
17.3 Preemptive Goal Programming
17.4 Summary
Glossary
Learning Aids for This Chapter
Problems
Case 17-1 A Cure for Cuba
Case 17-2 Remembering September 11
17.1
Chapter Eighteen
Inventory Management with Known Demand
18.1
18.2
18.3
18.4
18.5
18.6
A Case Study: The Atlantic Coast Tire Corp. (ACT)
Problem
Cost Components of Inventory Models
The Basic Economic Order Quantity (EOQ) Model
The Optimal Inventory Policy for the Basic EOQ
Model
The EOQ Model with Planned Shortages
The EOQ Model with Quantity Discounts
Chapter Ninteen
A Case Study for Perishable Products: Freddie the
Newsboy’s Problem
19.2 An Inventory Model for Perishable Products
19.3 A Case Study for Stable Products: The Niko Camera Corp. Problem
19.4 The Management Science Team’s Analysis of the
Case Study
19.5 A Continuous-Review Inventory Model for Stable
Products
19.6 Larger Inventory Systems in Practice
19.7 Summary
Glossary
Learning Aids for This Chapter
Problems
Case 19-1 TNT: Tackling Newsboy’s Teachings
Case 19-2 Jettisoning Surplus Stock
19.1
Chapter Twenty
Computer Simulation with Crystal Ball
A Case Study: Freddie the Newsboy’s Problem
Bidding for a Construction Project: A Prelude to the
Reliable Construction Co. Case Study
20.3 Project Management: Revisiting the Reliable Construction Co. Case Study
20.4 Cash Flow Management: Revisiting the Everglade
Golden Years Company Case Study
20.5 Financial Risk Analysis: Revisiting the Think-Big
Development Co. Problem
20.6 Revenue Management in the Travel Industry
20.7 Choosing the Right Distribution
20.8 Decision Making with Decision Tables
20.9 Optimizing with OptQuest
20.10 Summary
Glossary
Learning Aids for This Chapter
Solved Problem
Problems
Case 20-1 Action Adventures
Case 20-2 Pricing under Pressure
20.1
20.2
Chapter One
Introduction
Learning Objectives
After completing this chapter, you should be able to
1. Define the term management science.
2. Describe the nature of management science.
3. Explain what a mathematical model is.
4. Use a mathematical model to perform break-even analysis.
5. Use a spreadsheet model to perform break-even analysis.
6. Describe the relationship between analytics and management science.
7. Identify the levels of annual savings that management science sometimes can provide to
organizations.
8. Identify some special features of this book.
Welcome to the field of management science! We think that it is a particularly exciting and
interesting field. Exciting because management science is having a dramatic impact on the
profitability of numerous business firms around the world. Interesting because the methods
used to do this are so ingenious. We are looking forward to giving you a guided tour to introduce you to the special features of the field.
Some people approach management science with a certain amount of anxiety and skepticism. The main source of the anxiety is the reputation of the field as being highly mathematical. This reputation then generates skepticism that such a theoretical approach can have much
relevance for dealing with practical managerial problems. Most traditional courses (and textbooks) about management science have only reinforced these perceptions by emphasizing the
mathematics of the field rather than its practical application.
Rest easy. This is not a traditional management science textbook. We realize that most
readers of this book are aspiring to become managers, not mathematicians. Therefore, the
emphasis throughout is on conveying what a future manager needs to know about management science. Yes, this means including a little mathematics here and there, because it is a
major language of the field. The mathematics you do see will be at the level of high school
algebra plus (in the later chapters) basic concepts of elementary probability theory. We think
you will be pleasantly surprised by the new appreciation you gain for how useful and intuitive
mathematics at this level can be. However, managers do not need to know any of the heavy
mathematical theory that underlies the various techniques of management science. Therefore,
the use of mathematics plays only a strictly secondary role in the book.
One reason we can deemphasize mathematics is that powerful spreadsheet software is
available for applying management science. Spreadsheets provide a comfortable and familiar
environment for formulating and analyzing managerial problems. The spreadsheet takes care
of applying the necessary mathematics automatically in the background with only a minimum
of guidance by the user. This has revolutionized the use of management science. In the past,
technically trained management scientists were needed to carry out significant management
science studies for management. Now spreadsheets are bringing many of the tools and concepts of management science within the reach of managers for conducting their own analyses.
Although busy managers will continue to call upon management science teams to conduct
major studies for them, they are increasingly becoming direct users themselves through the
1
2 Chapter One Introduction
medium of spreadsheet software. Therefore, since this book is aimed at future managers (and
management consultants), we will emphasize the use of spreadsheets for applying management science.
What does an enlightened future manager need to learn from a management science course?
1. Gain an appreciation for the relevance and power of management science. (Therefore, we
include many application vignettes throughout the book that give examples of actual applications of management science and the impact they had on the organizations involved.)
2. Learn to recognize when management science can (and cannot) be fruitfully applied.
(Therefore, we will emphasize the kinds of problems to which the various management
science techniques can be applied.)
3. Learn how to apply the major techniques of management science to analyze a variety of managerial problems. (Therefore, we will focus largely on how spreadsheets enable many such
applications with no more background in management science than provided by this book.)
4. Develop an understanding of how to interpret the results of a management science study.
(Therefore, we will present many case studies that illustrate management science studies
and how their results depend on the assumptions and data that were used.)
The objectives just described are the key teaching goals of this book.
We begin this process in the next four sections by introducing the nature of management
science and the impact that it is having on many organizations. (This process will continue
throughout the remaining chapters as well.) Section 1.5 then points out some of the special
features of this book that you can look forward to seeing in the subsequent chapters.
1.1
THE NATURE OF MANAGEMENT SCIENCE
What is the name management science (sometimes abbreviated MS) supposed to convey? It
does involve management and science or, more precisely, the science of management, but this
still is too vague. Here is a more suggestive definition.
Management science is a discipline that attempts to aid managerial decision making by applying
a scientific approach to managerial problems that involve quantitative factors.
Now let us see how elaborating upon each of the italicized terms in this definition conveys
much more about the nature of management science.
Management Science Is a Discipline
As a discipline, management science is a whole body of knowledge and techniques that are
based on a scientific foundation. For example, it is analogous in some ways to the medical
field. A medical doctor has been trained in a whole body of knowledge and techniques that
are based on the scientific foundations of the medical field. After receiving this training and
entering practice, the doctor must diagnose a patient’s illness and then choose the appropriate
medical procedures to apply to the illness. The patient then makes the final decision on which
medical procedures to accept. For less serious cases, the patient may choose not to consult
a doctor and instead use his own basic knowledge of medical principles to treat himself.
Similarly, a management scientist must receive substantial training (albeit considerably less
than for a medical doctor). This training also is in a whole body of knowledge and techniques
that are based on the scientific foundations of the discipline. After entering practice, the
management scientist must diagnose a managerial problem and then choose the appropriate
management science techniques to apply in analyzing the problem. The cognizant manager
then makes the final decision as to which conclusions from this analysis to accept. For less
extensive managerial problems where management science can be helpful, the manager may
choose not to consult a management scientist and instead use his or her own basic knowledge
of management science principles to analyze the problem.
Although it has considerably longer roots, the rapid development of the discipline began in
the 1940s and 1950s. The initial impetus came early in World War II, when large numbers of
scientists were called upon to apply a scientific approach to the management of the war effort
1.1 The Nature of Management Science 3
operations research
Management science began
its rapid development during World War II with the
name operations research.
for the allies. When the war ended, the success of this approach in the war effort spurred interest in applying it outside the military as well. By the early 1950s, substantial applications of
management science were being seen in a variety of organizations in business, industry, and
government.
Another landmark event in the history of management science was the discovery in 1947
by George Dantzig of the simplex method for solving linear programming problems. (Linear programming is the subject of several early chapters.) Considerable progress in developing the other techniques of management science also occurred throughout the middle of the
20th century. However, the very limited computational power available at that time (whether
when doing the computations by hand or with the relatively primitive electronic computers of
the day) prevented applying these techniques except to small problems. Fortunately, another
factor that gave great impetus to the growth of the discipline ever since that time was the
onslaught of the computer revolution. Even massive problems usually can be solved now with
today’s powerful computers.
The traditional name given to the discipline (and the one that still is widely used today
outside of business schools) is operations research. This name was applied because the
teams of scientists in World War II were doing research on how to manage military operations. The abbreviation OR also is widely used. This abbreviation often is combined with the
one for management science (MS), thereby referring to the discipline as OR/MS. According to estimates from the U.S. Bureau of Labor Statistics in 2015, there were approximately
96,000 individuals at that time working as operations research analysts in the United States
(some with just a B.S. degree) with an average annual salary of about $84,000. The Bureau
also forecasted that this number of individuals working as operations research analysts would
grow by 30 percent over the subsequent decade.
Another discipline that is closely related to management science is analytics (sometimes
called business analytics when dealing with business problems). Like management science,
analytics attempts to aid managerial decision making but with particular emphasis on three
types of analysis: (1) descriptive analytics—the use of data (sometimes massive amounts of
data) to analyze trends, (2) predictive analytics—the use of data to predict what will happen in
the future (perhaps by using the forecasting techniques described in Chapter 10), and (3) prescriptive analytics—the use of data to prescribe the best course of action (frequently by using
the optimization techniques described throughout this book). Broadly speaking, the techniques
of the management science discipline provide the firepower for prescriptive analytics and, to
a lesser extent, for predictive analytics, but not so much for descriptive analytics. (Section 1.3
will further describe the relationship between analytics and management science.)
One major international professional society for the management science discipline (as
well as for business analytics) is the Institute for Operations Research and the Management
Sciences (INFORMS). Headquartered in the United States, with well over 12,000 members,
this society holds major conferences in the United States each year (including an annual
Conference for Business Analytics and Operations Research) plus occasional conferences
elsewhere. It also publishes several prominent journals, including Management Science,
Operations Research, Analytics, and Interfaces. (Articles describing actual applications of
management science are featured in Interfaces, so you will see many references and links to
this journal throughout the book.) In addition, a few dozen countries around the world have
their own national operations research societies. (More about this in Section 1.4.)
Thus, operations research/management science (OR/MS) is a truly international discipline.
(We hereafter will normally just use the name management science or the abbreviation MS.)
Management Science Aids Managerial Decision Making
The key word here is that management science aids managerial decision making. Management scientists don’t make managerial decisions. Managers do. A management science study
only provides an analysis and recommendations, based on the quantitative factors involved in
the problem, as input to the cognizant managers. Managers must also take into account various intangible considerations that are outside the realm of management science and then use
their best judgment to make the decision. Sometimes managers find that qualitative factors
are as important as quantitative factors in making a decision.
4 Chapter One Introduction
A small informal management science study might be conducted by just a single individual, who may be the cognizant manager. However, management science teams normally are
used for larger studies. (We often will use the term team to cover both cases throughout the
book.) Such a team often includes some members who are not management scientists but who
provide other types of expertise needed for the study. Although a management science team
often is entirely in-house (employees of the company), part or all of the team may instead
be consultants who have been hired for just the one study. Consulting firms that partially or
entirely specialize in management science currently are a growing industry.
Management Science Uses a Scientific Approach
Management science is based strongly on some scientific fields, including mathematics, statistics, and computer science. It also draws on the social sciences, especially economics. Since
the field is concerned with the practical management of organizations, a management scientist should have solid training in business administration, including its various functional
areas, as well.
To a considerable extent, a management science team will attempt to use the scientific
method in conducting its study. This means that the team will emphasize conducting a systematic investigation that includes careful data gathering, developing and testing hypotheses
about the problem (typically in the form of a mathematical model), and then applying sound
logic in the subsequent analysis.
When conducting this systematic investigation, the management science team typically
will follow the (overlapping) steps outlined and described below.
Step 1: Define the problem and gather data. In this step, the team consults with management to clearly identify the problem of concern and ascertain the appropriate objectives for
the study. The team then typically spends a surprisingly large amount of time gathering relevant data about the problem with the assistance of other key individuals in the organization.
A common frustration is that some key data are either very rough or completely unavailable.
This may necessitate installing a new computer-based management information system.
Another increasingly common problem is that there may be too much data available to
be easily analyzed. Dramatic advances in computerized data capture, processing power,
data transmission, and storage capabilities are enabling organizations to integrate their
various databases into massive data warehouses. This has led to the development of datamining software for extracting hidden predictive information, correlations, and patterns
from large databases.
Fortunately, the rapid development of the information technology (IT) field in recent
years is leading to a dramatic improvement in the quantity and quality of data that may be
available to the management science (MS) team. Corporate IT now is often able to provide
the computational resources and databases, as well as any helpful data mining, that are
needed by the MS team. Thus, the MS team often will collaborate closely with the IT group.
Step 2: Formulate a model (typically a mathematical model) to represent the
problem. Models, or approximate representations, are an integral part of everyday life.
Common examples include model airplanes, portraits, globes, and so on. Similarly, models play an important role in science and business, as illustrated by models of the atom,
models of genetic structure, mathematical equations describing physical laws of motion
or chemical reactions, graphs, organization charts, and industrial accounting systems.
Such models are invaluable for abstracting the essence of the subject of inquiry, showing
interrelationships, and facilitating analysis.
Mathematical models are also approximate representations, but they are expressed
in terms of mathematical symbols and expressions. Such laws of physics as F = ma and
E = mc2 are familiar examples. Similarly, the mathematical model of a business problem
is the system of equations and related mathematical expressions that describes the essence
of the problem.
With the emergence of powerful spreadsheet technology, spreadsheet models now
are widely used to analyze managerial problems. A spreadsheet model lays out the relevant data, measures of performance, interrelationships, and so forth, on a spreadsheet in
1.1 The Nature of Management Science 5
an organized way that facilitates fruitful analysis of the problem. It also frequently incorporates an underlying mathematical model to assist in the analysis, but the mathematics is
kept in the background so the user can concentrate on the analysis.
The modeling process is a creative one. When dealing with real managerial problems
(as opposed to some cut-and-dried textbook problems), there normally is no single “correct” model but rather a number of alternative ways to approach the problem. The modeling process also is typically an evolutionary process that begins with a simple “verbal
model” to define the essence of the problem and then gradually evolves into increasingly
more complete mathematical models (perhaps in a spreadsheet format).
We further describe and illustrate such mathematical models in the next section.
Step 3: Develop a computer-based procedure for deriving solutions to the problem
from the model. The beauty of a well-designed mathematical model is that it enables the
use of mathematical procedures to find good solutions to the problem. These procedures
usually are run on a computer because the calculations are too extensive to be done by
hand. In some cases, the management science team will need to develop the procedure. In
others, a standard software package already will be available for solving the model. When
the mathematical model is incorporated into a spreadsheet, the spreadsheet software normally includes a Solver that usually will solve the model.
Step 4: Test the model and refine it as needed. Now that the model can be solved, the
team needs to thoroughly check and test the model to make sure that it provides a sufficiently
accurate representation of the real problem. A number of questions should be addressed,
perhaps with the help of others who are particularly familiar with the problem. Have all the
relevant factors and interrelationships in the problem been accurately incorporated into the
model? Does the model seem to provide reasonable solutions? When it is applied to a past
situation, does the solution improve upon what was actually done? When assumptions about
costs and revenues are changed, do the solutions change in a plausible manner?
Step 5: Apply the model to analyze the problem and develop recommendations for
management. The management science team now is ready to solve the model, perhaps
under a variety of assumptions, in order to analyze the problem. The resulting recommendations then are presented to the managers who must make the decisions about how to
deal with the problem.
If the model is to be applied repeatedly to help guide decisions on an ongoing
basis, the team might also develop a decision support system. This is an interactive
­computer-based system that aids managerial decision making. The system draws current
data from databases or management information systems and then solves the various
­versions of the model specified by the manager.
Step 6: Help to implement the team’s recommendations that are adopted by
management. Once management makes its decisions, the management science team normally is asked to help oversee the implementation of the new procedures. This includes
providing some information to the operating management and personnel involved on the
rationale for the changes that are being made. The team also makes sure that the new
operating system is consistent with its recommendations as they have been modified and
approved by management. If successful, the new system may be used for years to come.
With this in mind, the team monitors the initial experience with the system and seeks to
identify any modifications that should be made in the future.
Management Science Considers Quantitative Factors
Many managerial problems revolve around such quantitative factors as production quantities,
revenues, costs, the amounts available of needed resources, and so on. By incorporating these
quantitative factors into a mathematical model and then applying mathematical procedures to
solve the model, management science provides a uniquely powerful way of analyzing such
managerial problems. Although management science is concerned with the practical management of organizations, including taking into account relevant qualitative factors, its special
contribution lies in this unique ability to deal with the quantitative factors.
6 Chapter One Introduction
The Special Products Company example discussed next will illustrate how management
science considers quantitative factors.
Review
Questions
1. When did the rapid development of the management science discipline begin?
2. What is the traditional name given to this discipline that still is widely used outside of business
schools?
3. What does a management science study provide to managers to aid their decision making?
4. Upon which scientific fields and social sciences is management science especially based?
5. What is a decision support system?
6. What are some common quantitative factors around which many managerial problems revolve?
1.2 AN ILLUSTRATION OF THE MANAGEMENT SCIENCE APPROACH:
BREAK-EVEN ANALYSIS
A cost that remains the
same regardless of the production volume is referred
to as a fixed cost, whereas
a cost that varies with the
production volume is called
a variable cost.
The Special Products Company produces expensive and unusual gifts to be sold in stores
that cater to affluent customers who already have everything. The latest new-product proposal to management from the company’s Research Department is a first-of-its-kind iWatch.
This iWatch would combine the features of a top-of-the-line atomic wristwatch and a next-­
generation smartphone, including the ability to respond to voice commands or questions with
voice responses. It also would connect to the Internet wirelessly to provide weather, sports
scores, stock quotes, and more. An extensive research-and-development project would be
needed to develop the iWatch. The proposal is to provide a generous budget of $10 million
for this project in order to provide as many desirable features as possible within this budget.
It is clear that the production costs for the iWatch would be very large because of the extreme
miniaturization that would be required, so the selling price would need to be far beyond the
reach of middle-class customers. Therefore, the marketing of the iWatch would be aimed at
wealthy customers who want the most advanced products regardless of cost.
Management needs to decide whether to develop and market this new product and, if so,
how many of these watches to produce. Before making these decisions, a sales forecast will
be obtained to estimate how many watches can be sold. Since most of these sales would
occur quickly during the relatively brief time before the “next big thing” arrives to take over
the market, there would be only one production run for the iWatch and the number produced
would be set equal to the sales forecast. Following the production run, the iWatch would be
marketed as aggressively as needed to sell this entire inventory if possible. Management now
needs a management science study to be conducted to determine how large this sales potential
needs to be to make the iWatch profitable after considering all the prospective revenues and
costs, so let’s next look at the estimates of these financial figures.
If the company goes ahead with this product, the research-and-development cost of
$10 million is referred to as a fixed cost because it remains the same regardless of how many
watches are produced and sold. (However, note that this cost would not be incurred if management decides not to introduce the product since the research-and-development project then
would not be undertaken.)
In addition to this fixed cost, there is a production cost that varies with the number of watches
produced. This variable cost is $1,000 per watch produced, which adds up to $1,000 times the
number of watches produced. (The cost for each additional unit produced, $1,000, is referred to
as the marginal cost.) Each watch sold would generate a unit revenue of $2,000 for the company.
Spreadsheet Modeling of the Problem
You will see throughout this book that spreadsheets provide a very convenient way of using a management science approach for modeling and analyzing a wide variety of managerial problems. This
certainly is true for the Special Products Company problem as well, as we now will demonstrate.
Figure 1.1 shows a spreadsheet formulation of this problem after obtaining a preliminary
sales forecast that indicates 30,000 watches can be sold. The data have been entered into cells
C4 to C7. Cell C9 is used to record a trial value for the decision as to how many watches to
produce. As one of the many possibilities that eventually might be tried, Figure 1.1 shows the
specific trial value of 20,000.
1.2 An Illustration of the Management Science Approach: Break-Even Analysis 7
FIGURE 1.1
A spreadsheet formulation of the Special Products Company problem.
A
1
B
C
Special Products Co. Break-Even Analysis
D
E
F
2
3
Data
4
5
Unit Revenue
Fixed Cost
$2,000
$10,000,000
6
Marginal Cost
7
Results
Total Revenue
Total Fixed Cost
$40,000,000
$10,000,000
$1,000
Total Variable Cost
$20,000,000
Sales Forecast
30,000
Profit (Loss)
$10,000,000
Production Quantity
20,000
8
9
Range Name
FixedCost
MarginalCost
ProductionQuantity
Profit
SalesForecast
TotalFixedCost
TotalRevenue
TotalVariableCost
UnitRevenue
Excel Tip: To update
formulas throughout the
spreadsheet to incorporate a
newly defined range name,
choose Apply Names from
the Define Name menu on
the Formulas tab.
Excel Tip: A list of all the
defined names and their
corresponding cell references can be pasted into a
spreadsheet by choosing
Paste Names from the Use
in Formula menu on the
Formulas tab, and then
clicking on Paste List.
A spreadsheet is a convenient tool for performing
what-if analysis.
The Excel function MIN
(a, b) gives the minimum
of the numbers in the cells
whose addresses are a
and b.
Cell
C5
C6
C9
F7
C7
F5
F4
F6
C4
E
3
4
5
6
7
F
Results
Total Revenue
=UnitRevenue * MIN(SalesForecast, ProductionQuantity)
Total Fixed Cost =IF(ProductionQuantity > 0, FixedCost, 0)
Total Variable Cost =MarginalCost * ProductionQuantity
Profit (Loss) =TotalRevenue – (TotalFixedCost + TotalVariableCost)
Cells F4 to F7 give the resulting total revenue, total costs, and profit (loss) by using the
Excel equations shown under the spreadsheet in Figure 1.1. The Excel equations could have
been written using cell references (e.g., F6 = C6*C9). However, the spreadsheet model
is made clearer by giving “range names” to key cells or blocks of cells. (A range name
is a descriptive name given to a cell or range of cells that immediately identifies what is
there. Appendix A provides details about how to incorporate range names into a spreadsheet model.) To define a name for a selected cell (or range of cells), click on the name box
(on the left of the formula bar above the spreadsheet) and type a name. These cell names
then can be used in other formulas to create an equation that is easy to decipher (e.g.,
TotalVariableCost = MarginalCost*ProductionQuantity rather than the more cryptic F6 =
C6*C9). Note that spaces are not allowed in range names. When a range name has more
than one word, we have used capital letters to distinguish the start of each new word (e.g.,
ProductionQuantity).
The lower left-hand corner of Figure 1.1 lists the names of the quantities in the spreadsheet
in alphabetical order and then gives cell references where the quantities are found. Although
this isn’t particularly necessary for such a small spreadsheet, you should find it helpful for the
larger spreadsheets found later in the book.
This same spreadsheet, along with all of the other spreadsheets in the book, are available
to you at www.mhhe.com/Hillier6e. As you can see for yourself by bringing up and playing
with the spreadsheet, it provides a straightforward way of performing what-if analysis on the
problem. What-if analysis involves addressing such questions as, What happens if the sales
forecast should have been considerably lower? What happens if some of the cost and revenue
estimates are wrong? Simply enter a variety of new values for these quantities in the spreadsheet and see what happens to the profit shown in cell F7.
The lower right-hand corner of Figure 1.1 introduces two useful Excel functions, the
MIN(a, b) function and the IF(a, b, c) function. The equation for cell F4 uses the MIN(a, b)
function, which gives the minimum of a and b. In this case, the estimated number of watches
that will be sold is the minimum of the sales forecast and the production quantity, so
​F4 = UnitRevenue*MIN(SalesForecast, ProductionQuantity)​
8 Chapter One Introduction
enters the unit revenue (from cell C4) times the minimum of the sales forecast (from C7) and
the production quantity (from C9) into cell F4.
Also note that the equation for cell F5 uses the IF(a, b, c) function, which does the following: If statement a is true, it uses b; otherwise, it uses c. Therefore,
The Excel function IF (a, b,
c) tests if a is true. If so, it
uses b; otherwise it uses c.
​F5 = IF(ProductionQuantity > 0, FixedCost, 0)​
says to enter the fixed cost (C5) into cell F5 if the production quantity (C9) is greater than
zero, but otherwise enter 0 (the fixed cost is avoided if production is not initiated).
The spreadsheet in Figure 1.1, along with its equations for the results in column F, constitutes a spreadsheet model for the Special Products Company problem. You will see many
examples of such spreadsheet models throughout the book.
This particular spreadsheet model is based on an underlying mathematical model that uses
algebra to spell out the equations in cells F4:F7 and then to derive some additional useful
information. Let us take a look at this mathematical model next.
Expressing the Problem Mathematically
The issue facing management is to make the following decision.
Decision to be made: Number of watches to produce (if any).
Since this number is not yet known, we introduce an algebraic variable Q to represent this
quantity. Thus,
​Q = Number of watches to produce,​
where Q is referred to as a decision variable. Naturally, the value chosen for Q should not
exceed the sales forecast for the number of watches that can be sold. Choosing a value of 0 for
Q would correspond to deciding not to introduce the product, in which case none of the costs
or revenues described in the preceding paragraph would be incurred.
The objective is to choose the value of Q that maximizes the company’s profit from this
new product. The management science approach is to formulate a mathematical model to
represent this problem by developing an equation that expresses the profit in terms of the
decision variable Q. To get there, it is necessary first to develop equations in terms of Q for
the total cost and revenue generated by the watches.
If Q = 0, no cost is incurred. However, if Q > 0, there is both a fixed cost and a variable cost.
Fixed cost = $10 million (if Q > 0)
​​
​  ​
​​​
Variable cost = $1,000Q
Therefore, the total cost would be
0
if Q = 0
​Total cost = ​​
​​​ ​
​
​​​​
{$10 million + $1,000Q
if Q > 0
Since each watch sold would generate a revenue of $2,000 for the company, the total revenue from selling Q watches would be
​Total revenue = $2,000Q​
Consequently, the profit from producing and selling Q watches would be
Profit = Total revenue − Total cost
if Q = 0​​
​ ​​
​  ​  0
​
​
= ​​ ​​​
​
​
​​​
{$2,000Q − ​​(​​$10 million + $1,000Q​)​​ ​ if Q > 0
Thus, since $2,000Q − $1,000Q = $1,000Q
​​Profit = −$10 million + $1,000Q​  if Q > 0​​
Analysis of the Problem
This last equation shows that the attractiveness of the proposed new product depends greatly
on the value of Q, that is, on the number of watches that can be produced and sold. A small
1.2 An Illustration of the Management Science Approach: Break-Even Analysis 9
FIGURE 1.2
Break-even analysis for
the Special Products
Company shows that the
cost line and revenue line
intersect at Q = 10,000
watches, so this is the
break-even point for the
proposed new product.
Revenue/Cost
in $millions
50
40
Profit
Total revenue = $2000Q
30
20
10
0
Total cost = $10 million + $1000Q if Q > 0
Fixed cost = $10 million if Q > 0
Loss
5
10
15
20
25
Q
Production Quantity (thousands)
Break-even point = 10,000 watches
value of Q means a loss (negative profit) for the company, whereas a sufficiently large value
would generate a positive profit for the company. For example, look at the difference between
Q = 2,000 and Q = 20,000.
Profit = − $10 million + $1,000 (2,000) = −$8 million
if Q = 2,000
​​
​  ​
​ ​​
Profit = − $10 million + $1,000 (20,000) = $10 million
if Q = 20,000
Figure 1.2 plots both the company’s total cost and total revenue for the various values
of Q. Note that the cost line and the revenue line intersect at Q = 10,000. For any value of
Q < 10,000, cost exceeds revenue, so the gap between the two lines represents the loss to the
company. For any Q > 10,000, revenue exceeds cost, so the gap between the two lines now
shows positive profit. At Q = 10,000, the profit is 0. Since 10,000 units is the production and
sales volume at which the company would break even on the proposed new product, this volume is referred to as the break-even point. This is the point that must be exceeded to make
it worthwhile to introduce the product. Therefore, the crucial question is whether the sales
forecast for how many watches can be sold is above or below the break-even point.
Figure 1.2 illustrates the graphical procedure for finding the break-even point. Another
alternative is to use an algebraic procedure to solve for the point. Because the profit is 0 at
this point, the procedure consists of solving the following equation for the unknown Q.
​​Profit = −$10 million + $1,000Q = 0​​
Thus,
$1,000Q = $10 million
$10 million
​​
​  =​ ​  __________
​​
​
Q​
​
​
$1,000
Q = 10,000
A Complete Mathematical Model for the Problem
The preceding analysis of the problem made use of a basic mathematical model that consisted
of the equation for profit expressed in terms of Q. However, implicit in this analysis were some
additional factors that can be incorporated into a complete mathematical model for the problem.
10 Chapter One Introduction
Two of these factors concern restrictions on the values of Q that can be considered. One of
these is that the number of watches produced cannot be less than 0. Therefore,
​Q ≥ 0​
constraints
A constraint in a mathematical model is an inequality
or equation that expresses
some restrictions on the
values that can be assigned
to the decision variables.
is one of the constraints for the complete mathematical model. Another restriction on the
value of Q is that it should not exceed the number of watches that can be sold. The preliminary sales forecast is 30,000, but this is such a crucial number that more time is needed now
to develop a final sales forecast that will be used hereafter. This final sales forecast has not yet
been obtained, so let the symbol s represent this currently unknown value.
​s = Sales forecast (not yet available) of the number of watches that can be sold​
Consequently,
​Q ≤ s​
parameter
The constants in a mathematical model are referred
to as the parameters of the
model.
is another constraint, where s is a parameter of the model whose value has not yet been chosen.
The final factor that should be made explicit in the model is the fact that management’s
objective is to make the decision that maximizes the company’s profit from this new product.
Therefore, the complete mathematical model for this problem is to find the value of the decision variable Q so as to
0
if Q = 0
​Maximize profit = ​​
​​​ ​
​
​​​​
{− $10 million + $1,000Q
if Q > 0
subject to
objective function
The objective function for
a mathematical model is a
mathematical expression
that gives the measure of
performance for the problem in terms of the decision
variables.
Q≤s
​​
​​​
Q≥0
where the algebraic expression given for Profit is called the objective function for the
model. The value of Q that solves this model depends on the value that will be assigned to
the parameter s (the future final forecast of the number of units that can be sold). Because the
break-even point is 10,000, here is how the solution for Q depends on s.
Solution for Mathematical Model:
Fixed cost
$10 million
_________________________
Break-even
point = ​
​ = ______________
​
​​
​​
​ ​ − $1,000 ​= 10,000​
Unit
revenue
− Marginal cost $2,000
If s ≤ 10,000, then set Q = 0.
If s >10,000, then set Q = s.
Therefore, the company should introduce the product and produce the number of units that
can be sold only if this production and sales volume exceeds the break-even point.
What-if Analysis of the Mathematical Model
what-if analysis
Since estimates can be
wrong, what-if analysis is
used to check the effect on
the recommendations of a
model if the estimates turn
out to be wrong.
A mathematical model is intended to be only an approximate representation of the problem.
For example, some of the numbers in the model inevitably are only estimates of quantities that
cannot be determined precisely at this time.
The above mathematical model is based on four numbers that are only estimates—the
fixed cost of $10 million, the marginal cost of $1,000, the unit revenue of $2,000, and the final
sales forecast (after it is obtained). A management science study usually devotes considerable time to investigating what happens to the recommendations of the model if any of the
estimates turn out to considerably miss their targets. This is referred to as what-if analysis.
Incorporating the Break-Even Point into the Spreadsheet Model
A key finding of the above mathematical model is its formula for the break-even point,
Fixed cost
​Break-even point = _________________________
​      ​​
Unit revenue − Marginal cost
Therefore, once both the quantities in this formula and the sales forecast have been carefully
estimated, the solution for the mathematical model specifies what the production quantity
should be.
By contrast, although the spreadsheet in Figure 1.1 enables trying a variety of trial
values for the production quantity, it does not directly indicate what the production quantity
1.2 An Illustration of the Management Science Approach: Break-Even Analysis 11
FIGURE 1.3
An expansion of the spreadsheet in Figure 1.1 that uses the solution for the mathematical model to calculate the break-even point.
A
1
B
C
Special Products Co. Break-Even Analysis
D
E
F
2
3
Data
4
5
Unit Revenue
Fixed Cost
$2,000
$10,000,000
6
Marginal Cost
7
Results
Total Revenue
Total Fixed Cost
$60,000,000
$10,000,000
$1,000
Total Variable Cost
$30,000,000
Sales Forecast
30,000
Profit (Loss)
$20,000,000
Production Quantity
30,000
Break-Even Point
8
9
Range Name
BreakEvenPoint
FixedCost
MarginalCost
ProductionQuantity
Profit
SalesForecast
TotalFixedCost
TotalRevenue
TotalVariableCost
UnitRevenue
Cell
F9
C5
C6
C9
F7
C7
F5
F4
F6
C4
E
3
4
5
6
7
10,000
F
Results
Total Revenue
=UnitRevenue * MIN(SalesForecast, ProductionQuantity)
Total Fixed Cost =IF(ProductionQuantity > 0, FixedCost, 0)
Total Variable Cost =MarginalCost * ProductionQuantity
Profit (Loss) =TotalRevenue – (TotalFixedCost + TotalVariableCost)
8
9
Break-Even Point
=FixedCost/(UnitRevenue – MarginalCost)
should be. Figure 1.3 shows how this spreadsheet can be expanded to provide this additional
guidance.
Suppose that a final sales forecast of 30,000 (unchanged from the preliminary sales forecast) now has been obtained, as shown in cell C7. As indicated by its equation at the bottom of
the figure, cell F9 calculates the break-even point by dividing the fixed cost ($10 million) by
the net profit per watch sold ($1,000), where this net profit is the unit revenue ($2,000) minus
the marginal cost ($1,000). Since the sales forecast of 30,000 exceeds the break-even point of
10,000, this forecast has been entered into cell C9.
If desired, the complete mathematical model for break-even analysis can be fully incorporated into the spreadsheet by requiring that the model solution for the production quantity be
entered into cell C9. This would be done by using the equation
​C9 = IF(SalesForecast > BreakEvenPoint, SalesForecast, 0)​
However, the disadvantage of introducing this equation is that it would eliminate the possibility of trying other production quantities that might still be of interest. For example, if
management does not have much confidence in the final sales forecast and wants to minimize
the danger of producing more watches than can be sold, consideration would be given to
production quantities smaller than the forecast. For example, the trial value shown in cell C9
of Figure 1.1 might be chosen instead. As in any application of management science, a mathematical model can provide useful guidance but management needs to make the final decision
after considering factors that may not be included in the model.
Review
Questions
1. How do the production and sales volume of a new product need to compare to its break-even
point to make it worthwhile to introduce the product?
2. What are the factors included in the complete mathematical model for the Special Products
Company problem, in addition to an equation for profit?
3. What is the purpose of what-if analysis?
4. How can a spreadsheet be used to perform what-if analysis?
5. What does the MIN(a, b) Excel function do?
6. What does the IF(a, b, c) Excel function do?
12 Chapter One Introduction
1.3 THE RELATIONSHIP BETWEEN ANALYTICS AND
MANAGEMENT SCIENCE
Analytics is a broad term
that includes both management science and all the
other quantitative decision
sciences, such as mathematics, statistics, computer
science, data science,
industrial engineering, etc.
By drawing on any of these
tools to analyze the available data, analytics can be
defined as the scientific
process of transforming
data into insight for making
better decisions.
Analytics encompasses all
of the quantitative decision
sciences.
The era of big data has created new challenges that
require the use of analytics.
There has been great buzz throughout the business world in recent years about something called
analytics (or business analytics) and the importance of incorporating analytics into managerial decision making. The primary impetus for this buzz was a series of articles and books
by Thomas H. Davenport, a renowned thought-leader who has helped hundreds of companies
worldwide to revitalize their business practices. He initially introduced the concept of analytics
in the January 2006 issue of the Harvard Business Review with an article, “Competing on Analytics,” that now has been named as one of the ten must-read articles in that magazine’s 90-year
history. This article soon was followed by two best-selling books entitled Competing on Analytics: The New Science of Winning and Analytics at Work: Smarter Decisions, Better Results.
So what is analytics? In contrast to management science, analytics is not a single discipline
with its own well-defined body of techniques. Analytics instead includes many disciplines,
namely, all the quantitative decision sciences.
Thus, any application of analytics draws on any of the quantitative decision sciences that
can be most helpful in analyzing a given problem. Therefore, a company’s analytics group
might include members with titles such as mathematician, statistician, computer scientist,
data scientist, information technologist, business analyst, industrial engineer, management
scientist, and operations research analyst.
At this point, the members of a management science group might object, saying that their
management science studies often draw upon these other quantitative decision sciences as
well. This frequently is true, but it also is true that some applications of analytics draw mainly
from certain other quantitative decision sciences instead of management science. This typically occurs when the issue being addressed is to try to gain insights from massive amounts
of available data, so that data science and statistics become the key quantitative decision sciences. This kind of application is a major emphasis of analytics.
The reason for this emphasis is that analytics fully recognizes that we have entered into the
era of big data where massive amounts of data now are commonly available to many businesses and organizations to help guide managerial decision making. The current data surge is
coming from sophisticated computer tracking of shipments, sales, suppliers, and customers,
as well as e-mail, web traffic, and social networks. A primary focus of analytics is on how to
make the most effective use of all these data.
The application of analytics can be divided into three overlapping categories. Here are the
traditional names and brief definitions of these categories:
Category 1: Descriptive analytics (analyzing data to improve descriptions of what has
been happening)
Category 2: Predictive analytics (using predictive models to improve predictions of what
is likely to happen in the future)
Category 3: Prescriptive analytics (using prescriptive models, including optimization
models, to improve managerial decision making)
The first of these categories requires dealing with perhaps massive amounts of historical
data where information technology has been used to store and access the data. Descriptive
analytics then uses innovative techniques to locate the relevant data and identify the interesting
patterns in order to describe and understand what is going on now. One important technique
for doing this is called data mining. A software package accompanying this book, Analytic
Solver, includes a tab called Data Mining for applying both data mining and predictive analytics. Because the analysis of data is the focus of the work being done in this category, data analytics is another name that is sometimes used for this category (and perhaps category 2 as well).
Some analytics professionals who specialize in descriptive analytics are called data scientists.
Predictive analytics involves applying predictive models to historical data and perhaps
external data to predict future events or trends. The models underlying statistical forecasting methods, such as those described in Chapter 10, are often used here. Computer simulation (Chapters 12 and 13) also can be useful for demonstrating future events that can occur.
Because some of the methods of predictive analytics are relatively sophisticated, this category
tends to be more advanced than the first one.
1.3 The Relationship Between Analytics and Management Science 13
Prescriptive analytics uses
powerful techniques drawn
mainly from management
science to prescribe what
should be done in the
future.
Sports analytics is another
example of various areas
where analytics now is
applied.
Business schools now are
responding to the great
need to thoroughly train
much larger numbers
of people in business
analytics.
Prescriptive analytics is the final (and most advanced) category. It involves applying
sophisticated models to the data to prescribe what should be done in the future. The powerful optimization models and techniques of management science described in many of the
chapters of this book commonly are what are used here. The purpose is to guide managerial
decision making, so the name decision analytics also could be used to describe this category.
Management scientists often deal with all three of these categories, but not very much
with the first one, somewhat more with the second one, and then heavily with the third one.
Thus, management science can be thought of as focusing mainly on advanced analytics (predictive and prescriptive activities) whereas analytics professionals might get more involved
than management scientists with the entire business process, including what precedes the first
category (identifying a need) and what follows the last category (implementation). Looking to
the future, the two approaches should tend to merge somewhat over time.
Although analytics was initially introduced as a key tool for mainly business organizations,
it also can be a powerful tool in other contexts. As one example, analytics (together with management science) played a key role in the 2012 presidential campaign in the United States.
The Obama campaign management hired a multi-disciplinary team of statisticians, predictive
modelers, data-mining experts, mathematicians, software programmers, and management scientists. It eventually built an entire analytics department five times as large as that of its 2008
campaign. With all this analytics input, the Obama team launched a full-scale and all-front
campaign, leveraging massive amounts of data from various sources to directly micro-target
potential voters and donors with tailored messages. The election had been expected to be a
very close one, but the Obama “ground game” that had been propelled by descriptive and
predictive analytics was given much of the credit for the clear-cut Obama win.
Another famous application of analytics is described in the book Moneyball and a subsequent
2011 movie with the same name that is based on this book. They tell the true story of how the
Oakland Athletics baseball team achieved great success, despite having one of the smallest budgets
in the major leagues, by using various kinds of nontraditional data (referred to as saber metrics)
to better evaluate the potential of players available through a trade or the draft. Although these
evaluations often flew in the face of conventional baseball wisdom, both descriptive analytics and
predictive analytics were being used to identify overlooked players who could greatly help the
team. After witnessing the impact of analytics, many major league baseball teams now have hired
analytics professionals. Some other kinds of sports teams, including professional basketball teams
in the NBA (National Basketball Association) also have begun to use analytics. For example, some
months after they won the 2015 NBA championship, the Golden State Warriors won the “Best
Analytics Organization” award at the MIT Sloan Sports Analytics Conference in March 2016.
The top management of numerous business organizations now understand the impact that this
approach can have on the bottom line and they are interested in increasing the role of their analytics group in their organization. This will require many more people trained in analytics and management science. A recent study by the McKinsey Global Institute estimated that the United States
could face a shortage by 2018 of 140,000 to 180,000 people with deep analytical skills. The study
also estimated that employers will need an additional 1.5 million managers and analysts with the
experience and expertise to use the analysis of big data to make effective and efficient decisions.
Universities now are responding to this great need. There now are hundreds of business
schools in the United States or abroad that have, or have committed to launch, curriculum
at the undergraduate and graduate levels with degrees or certificates in business analytics.
Courses that cover the material in this book would be a key component of these programs,
along with courses that emphasize other areas of analytics (e.g., statistics and data mining).
This creates an outstanding opportunity for students with a STEM focus. In the words of the
thought leader Thomas H. Davenport (who was introduced in the first paragraph of this section), the job of an analytics professional promises to be the “sexiest job in the 21st century.”
In 2016, the job site Glassdoor also named data scientist as the best job in America. Along
these same lines, U.S. News annually publishes a list of the best jobs in the United States,
based on a variety of factors such as salary, job satisfaction, etc. In their issue of January
27, 2016, U.S. News published that year’s list that placed the operations research profession
(i.e., the management science profession) as number 2 on the list of the best business jobs in
the country. Furthermore, this profession ranks very high in terms of having a high percentage
of women working in the profession, with slightly more women than men in the field.
14 Chapter One Introduction
Management science is
at the core of advanced
analytics.
Review
Questions
1.4
The momentum of the analytics movement is indeed continuing to grow rapidly. Because
management science is at the core of advanced analytics, the usage of the powerful techniques
of management science introduced in this book also should continue to grow rapidly. However, without even looking to the future, the impact of management science over past years
also has been impressive, as described in the next section.
1.
2.
3.
4.
5.
6.
What are some quantitative decision sciences that are included within analytics?
What does descriptive analytics involve doing?
What does predictive analytics involve doing?
What does prescriptive analytics involve doing?
In which categories of analytics does management science mainly focus?
What is a major emphasis of analytics that draws mainly from certain other quantitative decision sciences instead of management science?
THE IMPACT OF MANAGEMENT SCIENCE
The most important applications of management
science in business and
industry have resulted in
annual savings in the hundreds of millions of dollars.
Management science also
has had a major impact
in the health care area, in
guiding key governmental
policies, and in military
applications.
Management science (or operations research as it is commonly called by practitioners) has
had an impressive impact on improving the efficiency of numerous organizations around the
world. In the process, management science has made a significant contribution to increasing
the productivity of the economies of various countries. There now are a few dozen member countries in the International Federation of Operational Research Societies (IFORS),
with each country having a national operations research society. Both Europe and Asia have
federations of such societies to coordinate holding international conferences and publishing
international journals in those continents. In addition, we described in Section 1.1 how the
Institute for Operations Research and the Management Sciences (INFORMS) is a particularly prominent international society in this area. Among its various journals is one called
­Interfaces that regularly publishes articles describing major management science studies and
the impact they had on their organizations.
Management science (MS) has had numerous applications of various types in business and
industry, sometimes resulting in annual savings of millions, or even hundreds of millions, of
dollars. As an example, many hundreds of management scientists work on such airline problems as how to most effectively assign airplanes and crews to flights and how to develop fare
structures that maximize revenue. For decades, financial services firms have used portfolio
selection techniques that were developed by management scientists who won the Nobel Prize
in Economics for their work. Management science models have become a core component of
the marketing discipline. Multinational corporations rely on MS for guiding the management
of their supply chains. There are numerous other examples of MS applications that are having
a dramatic impact on the companies involved.
Management science also is widely used outside business and industry. For example, it
is having an increasing impact in the health care area, with applications involving improved
management of health care delivery and operations, disease modeling, clinical diagnosis and
decision making, radiation therapy, and so on. Applications of MS also abound at various
levels of government, ranging from dealing with national security issues at the federal level to
managing the delivery of emergency services at the municipal level. Other key governmental
applications involve the use of MS modeling to help guide energy, environmental, and global
warming policies. Some of the earliest MS applications were military applications, including
logistical planning and war gaming, and these continue today.
These are just a sampling of the numerous applications of management science that are
having a major impact on the organizations involved. The list goes on and on.
To give you a better notion of the wide applicability of management science, we list some
actual applications in Table 1.1. Note the diversity of organizations and applications in the first
An Application Vignette
General Motors (GM) is one of the largest and most successful
companies in the world. One major reason for this great success is
that GM also is one of the world’s leading users of advanced analytics and management science. In recognition of the great impact
that the application of these techniques has had on the success of
the company, GM was awarded the 2016 INFORMS Prize.
INFORMS (The Institute for Operations Research and the
Management Sciences) awards the INFORMS Prize to just one
organization each year for its particularly exceptional record
in applying advanced analytics and operations research/­
management science (OR/MS) throughout the organization.
The award winner must have repeatedly applied these techniques in pioneering, varied, novel, and lasting ways. Following
is the citation that describes why GM won the prize for 2016.
Citation: The 2016 INFORMS Prize is awarded to General
Motors for its sustained record of innovative and impactful
applied operations research and advanced analytics.
General Motors has hundreds of OR/MS practitioners worldwide who play a vital role in driving data-driven decisions in
everything from designing, building, selling, and servicing
vehicles to purchasing, logistics and quality. The team is constantly developing new business models and vetting emerging
opportunities.
GM has developed new market research and analysis techniques to understand what products and features customers
most want, to determine the ideal vehicles for their dealers to
stock, and to identify the steps they can take to achieve GM’s
goal of creating customers for life.
GM is also leading the industry by using data science and
advanced analytics to predict failure of automotive components and systems before customers are inconvenienced. GM’s
­industry-first Proactive Alert messages notify customers through
their OnStar system of a possible malfunction, transforming a
potential emergency repair into routine planned maintenance.
“Over the last seven decades, OR/MS techniques have been
used to improve our understanding of everything from traffic science and supply chain logistics to manufacturing productivity, product development, vehicles telematics and prognostics,” said Gary
Smyth, executive director of GM Global R&D Laboratories. “These
approaches to problem solving permeate almost everything we do.”
The impact OR/MS is now having on its business accelerated
in 2007, when GM created a center of expertise for Operations
Research to promote best practices and transfer new technologies. It since has expanded to include partner teams in product
development, supply chain, finance, information technology and
other functions.
(Note: The application vignette in Section 11.5 describes just
one example of the numerous applications of advanced analytics and management science at GM, where this one led to
­billions of dollars in savings and increased revenue.)
“Past INFORMS Awards, 2016: General Motors,” INFORMS.com
©INFORMS. Used with permission. All rights reserved.
TABLE 1.1 Applications of Management Science to Be Described in Application Vignettes
Organization
Area of Application
Section
Annual Savings
General Motors
Swift & Company
Samsung Electronics
Numerous applications
Improve sales and manufacturing performance
Reduce manufacturing times and inventory levels
1.4
2.1
2.7
INDEVAL
Chevron
Taylor Communications
Welch’s
Hewlett-Packard
Norwegian companies
Canadian Pacific Railway
Waste Management
MISO
Netherlands Railways
Continental Airlines
Bank Hapoalim Group
Settle all securities transactions in Mexico
Optimize refinery operations
Assign print jobs to printers
Optimize use and movement of raw materials
Product portfolio management
Maximize flow of natural gas through offshore pipeline network
Plan routing of rail freight
Develop a route-management system for trash collection and disposal
Administer the transmission of electricity in 13 states
Optimize operation of a railway network
Reassign crews to flights when schedule disruptions occur
Develop a decision-support system for investment advisors
3.2
3.4
3.6
4.3
6.1
6.3
6.4
7.1
7.2
7.4
7.5
8.2
DHL
Workers’ Compensation Board
CDC
L. L. Bean
Taco Bell
CSAV
General Motors
Optimize the use of marketing resources
Manage high-risk disability claims and rehabilitation
Eradicate polio
Forecast staffing needs at call centers
Forecast the level of business throughout the day
Optimize global shipping
Improve the throughput of its production lines
8.4
9.3
9.7
10.2
10.3
10.6
11.5
Not estimated
$12 million
$200 million
more revenue
$150 million
Nearly $1 billion
$10 million
$150,000
$180 million
$140 million
$100 million
$100 million
$700 million
$105 million
$40 million
$31 million
more revenue
$260 million
$4 million
$1.5 billion
$300,000
$13 million
$81 million
$150 million
Federal Aviation Administration
Sasol
Merrill Lynch
Manage air traffic flows in severe weather
Improve the efficiency of its production processes
Pricing analysis for providing financial services
12.2
12.3
13.4
Kroger
Pharmacy inventory management
13.8
$200 million
$23 million
$50 million
more revenue
$10 million
15
16 Chapter One Introduction
two columns. The third column identifies the section where an “application vignette” devotes
several paragraphs to describing the application and also references an article that provides full
details. (This section includes the first of these application vignettes, where this one mentions a
wide variety of important applications.) The last column indicates that these applications typically resulted in annual savings in the many millions of dollars. Furthermore, additional benefits not recorded in the table (e.g., improved service to customers and better managerial control)
sometimes were considered to be even more important than these financial benefits. (You will
have an opportunity to investigate these less tangible benefits further in Problems 1.9 and 1.10.)
A link to the articles in Interfaces that describe these applications in detail is provided at
www.mhhe.com/Hillier6e. We are grateful to INFORMS for our special partnership to make
these articles available to you through this link. We think you will find these articles interesting and enlightening in illustrating the dramatic impact that management science sometimes
can have on the success of a variety of organizations.
You also will see a great variety of applications of management science throughout the
book in the form of case studies, examples, and end-of-chapter cases. Some of the applications are similar to ones described in application vignettes, but many others are quite different. However, they all generally fall into one of three broad categories, namely, applications
in the areas of operations management, finance, and marketing. Tables 1.2, 1.3, and 1.4 list
TABLE 1.2
Case Studies and
­Examples in the Area of
Operations Management
Location
Type of Application
Sec. 2.1 onward
Case 2-1
Case 2-2
Case 2-3
Sec. 3.3
Sec. 3.5
Sec. 3.6
Case 3-1
Case 3-3
Cases 3-5, 5-4, 7-3
Case 3-6
A case study: What is the most profitable mix of products?
Which mix of car models should be produced?
Which mix of ingredients should go into the casserole in a university cafeteria?
Which mix of customer–service agents should be hired to staff a call center?
Personnel scheduling of customer–service agents
Minimize the cost of shipping a product from factories to customers
Optimize the assignment of personnel to tasks
How should a product be shipped to market?
Which mix of women’s clothing should be produced for next season?
Develop a plan for assigning students to schools so as to minimize busing costs
Which mixes of solid waste materials should be amalgamated into different
grades of a salable product?
How should qualified managers be assigned to new R&D projects?
Develop and analyze a steel company’s plan for pollution abatement
Plan the mix of livestock and crops on a farm with unpredictable weather
Minimize the cost of operating a distribution network
A case study: Maximize the flow of goods through a distribution network
Find the shortest path from an origin to a destination
Logistical planning for a military campaign
Develop the most profitable flight schedules for an airline
Operate and expand a private computer network
Choose the best combination of R&D projects to pursue
Select the best sites for emergency services facilities
Airline crew scheduling
Production planning when setup costs are involved
Make inventory decisions for a retailer’s warehouse
Production planning when overtime is needed
Find the shortest route to visit all the American League ballparks
Many examples of commercial service systems, internal service systems, and
transportation service systems that can be analyzed with queueing models
A case study: An analysis of competing proposals for more quickly providing
maintenance services to customers
Analysis of proposals for reducing in-process inventory
Comparison of whether corrective maintenance or preventive maintenance is better
A case study: Would it be profitable for the owner of a small business to add an
associate?
Analysis of proposals for relieving a production bottleneck
A case study: How much of a perishable product should be added to a retailer’s
inventory?
Plan a complex project to ensure a strong likelihood of meeting the project deadline
Case 3-7
Case 5-2
Case 5-3
Sec. 6.1
Secs. 6.2, 6.3
Sec. 6.4
Case 6-1
Case 6-3
Cases 6-4 , 7-4
Sec. 7.2
Sec. 7.3
Sec. 7.4
Sec. 7.5
Case 7-2
Sec. 8.3
Sec. 8.5
Sec. 11.2
Sec. 11.4 onward
Cases 11-2, 12-2
Sec. 12.1
Secs. 12.2, 12.3
Case 12-1
Sec. 13.1
Sec. 13.3
1.4 The Impact of Management Science 17
TABLE 1.3
Case Studies and
Examples in the Area of
Finance
TABLE 1.4
Case Studies and
Examples in the Area of
Marketing
Location
Type of Application
Sec. 1.2
Case 1-1
Sec. 3.2
Sec. 3.2
Case 3-2
Sec. 4.1 onward
Case 4-1
Sec. 6.4
Case 6-2
Sec. 7.1
Case 7-1
Sec. 8.2
Sec. 8.5
Case 8-2
Case 8-3
Sec. 9.1 onward
Case 9-1
Sec. 12.1
Sec. 13.2
Sec. 13.4
Sec. 13.5
Sec. 13.6
Case 13-1
Case 13-2
Break-even analysis
Break-even analysis and what-if analysis
An airline choosing which airplanes to purchase
Capital budgeting of real-estate development projects
Develop a schedule for investing in a company’s computer equipment
A case study: Develop a financial plan for meeting future cash flow needs
Develop an investment and cash flow plan for a pension fund
Minimize the cost of car ownership
Find the most cost-effective method of converting various foreign currencies into
dollars
A case study: Determine the most profitable combination of investments
Develop an investment plan for purchasing art
Portfolio selection that balances expected return and risk
Select a portfolio to beat the market as frequently as possible
Determine an optimal investment portfolio of stocks
Develop a long-range plan to purchase and sell international bonds
A case study: Choose whether to drill for oil or sell the land instead
Choose a strategy for the game show, “Who Wants to Be a Millionaire?”
Analysis of a new gambling game
Choose the bid to submit in a competitive bidding process
Develop a financial plan when future cash flows are somewhat unpredictable
Risk analysis when assessing financial investments
How much overbooking should be done in the travel industry?
Analysis of how a company’s cash flows might evolve over the next year
Calculate the value of a European call option
Location
Type of Application
Secs. 2.7, 3.3
Case 2-1
Secs. 3.1, 3.4
Case 3-4
Case 5-1
Determine the best mix of advertising media
Evaluate whether an advertising campaign would be worthwhile
A case study: Which advertising plan best achieves managerial goals?
Develop a representative marketing survey
Analysis of the trade-off between advertising costs and the resulting increase in
sales of several products
Balance the speed of bringing a new product to market and the associated costs
Deal with nonlinear marketing costs
Refine the advertising plan developed in the case study presented in Sections 3.1
and 3.4
Should a company immediately launch a new product or test-market it first?
Should a company buy additional marketing research before deciding whether to
launch a new product?
Plan a sequence of decisions for a possible new product
A case study: Manage a call center for marketing goods over the telephone
Improve forecasts of demand for a call center
Estimate customer waiting times for calling into a call center
Sec. 6.4
Sec. 8.3
Case 8-1
Case 9-2
Case 9-3
Case 9-4
Sec. 10.2 onward
Case 10-1
Case 11-1
these applications in these three respective areas, where the first column identifies where
the application is described. In the second column of each table, note the many different
ways in which management science can have a real impact in helping improve managerial
decisions.
Even these long lists of applications in Tables 1.1 to 1.4 are just a sample of the numerous important ways in which management science is applied in organizations around the
world. We do not have enough space to provide a more comprehensive compilation of the
important applications. (Other applications are included in the supplementary chapters at
www.mhhe.com/Hillier6e.) A hallmark of management science is its great flexibility in dealing with new managerial problems as they arise.
18 Chapter One Introduction
1.5
SOME SPECIAL FEATURES OF THIS BOOK
The focus of this book is on teaching what an enlightened future manager needs to learn from
a management science course. It is not on trying to train technical analysts. This focus has led
us to include a number of special features that we hope you enjoy.
One special feature is that the entire book revolves around modeling as an aid to managerial decision making. This is what is particularly relevant to a manager. Although they may
not use this term, all managers often engage in at least informal modeling (abstracting the
essence of a problem to better analyze it), so learning more about the art of modeling is
important. Since managers instigate larger management science studies done by others, they
also need to be able to recognize the kinds of managerial problems where such a study might
be helpful. Thus, a future manager should acquire the ability both to recognize when a management science model might be applicable and to properly interpret the results from analyzing the model. Therefore, rather than spending substantial time in this book on mathematical
theory, the mechanics of solution procedures, or the manipulation of spreadsheets, the focus is
on the art of model formulation, the role of a model, and the analysis of model results. A wide
range of model types is considered.
Another special feature is a heavy emphasis on case studies to better convey these ideas in
an interesting way in the context of applications. Every subsequent chapter includes at least
one case study that introduces and illustrates the application of that chapter’s techniques in a
realistic setting. In a few instances, the entire chapter revolves around a case study. Although
considerably smaller and simpler than most real studies (to maintain clarity), these case studies are patterned after actual applications requiring a major management science study. Consequently, they convey the whole process of such a study, some of the pitfalls involved, and
the complementary roles of the management science team and the manager responsible for
the decisions to be made.
To complement these case studies, every chapter also includes major cases at the end.
These realistic cases can be used for individual assignments, team projects, or case studies in
class. In addition, the University of Western Ontario Ivey School of Business (the second largest producer of teaching cases in the world) also has specially selected cases from its case collection that match the chapters in this textbook. These cases are available on the Ivey Website,
www.cases.ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book.
The book also places heavy emphasis on conveying the impressive impact that management
science is having on improving the efficiency of numerous organizations around the world.
Therefore, you will see many examples of actual applications throughout the book in the form of
boxed application vignettes. You then will have the opportunity to learn more about these actual
applications by reading the articles fully describing them through using the links provided in
www.mhhe.com/Hillier6e. As indicated in Table 1.1, these applications sometimes resulted in
annual savings of millions, tens of millions, or even hundreds of millions of dollars.
In addition, we try to provide you with a broad perspective about the nature of the real
world of management science in practice. It is easy to lose sight of this world when cranking
through textbook exercises to master the mechanics of a series of techniques. Therefore, we
shift some emphasis from mastering these mechanics to seeing the big picture. The case studies, cases, and descriptions of actual applications are part of this effort.
A new feature with this edition is the inclusion of a McGraw-Hill web-based learning
tool called Connect. In addition to providing various instructor resources, Connect includes
a powerful tool called SmartBook that provides personalized instruction to help students
better learn the material. Many students might benefit by using SmartBook. This tool can be
accessed at connect.mheducation.com. The preface provides more information about both
Connect and SmartBook.
Another feature is the inclusion of one or more solved problems for each chapter to help
you get started on your homework for that chapter. The statement of each solved problem is
given just above the Problems section of the chapter, and then you can find the complete solutions at www.mhhe.com/Hillier6e.
As described further in the preface, this same website provides a full range of resources
of interest to students, including access to supplementary text material (both supplements to
1.5 Some Special Features of this Book 19
book chapters and additional chapters) and much more. For example, this website includes
numerous spreadsheet files for every chapter in this book. Each time a spreadsheet example
is presented in the book, a live spreadsheet that shows the formulation and solution for the
example also is provided. This gives a convenient reference, or even useful templates, when
you set up spreadsheets to solve similar problems. Also, for many models in the book, template spreadsheet files are provided that already include all the equations necessary to solve
the model. You simply enter the data for the model and the solution is immediately calculated.
The last, but certainly not the least, of the special features of this book is the accompanying software. We will describe and illustrate how to use today’s premier spreadsheet package,
Microsoft Excel, to formulate many management science models in a spreadsheet format. Many
of the models considered in this book can be solved using standard Excel. Some Excel add-ins
also are available to solve other models. Appendix A provides a primer on the use of Excel.
Included with standard Excel is an add-in, called Solver, which is used to solve most of
the optimization models considered in the first half of this book. Also included with this
edition of the textbook is a very powerful software package from Frontline Systems, Inc.,
called Analytic Solver® for Education (hereafter referred to as Analytic Solver). Instructions
for downloading this software are available at www.mhhe.com/Hillier6e. Some special features of Analytic Solver are a significantly enhanced version of the basic Solver included with
Excel, the ability to build decision trees within Excel, as covered in Chapter 9, forecasting
tools, as covered in Chapter 10, and tools to build computer simulation models within Excel,
as covered in Chapter 13. In addition to the Analytic Solver add-in for Excel, you also have
access to AnalyticSolver.com. This is cloud-based software accessed through a browser. It has
the same look and essentially all the features of Analytic Solver in Excel, but can be used by
any computer with access to the Internet without downloading additional software.
Most of the software used in this book is compatible with both Excel for Windows PCs
and Excel for Macintosh computers (Macs). Analytic Solver is not directly compatible with
Macs, although it works well on any recent Mac with Boot Camp or virtualization software.
AnalyticSolver.com works through a browser, so can be used with both Windows and Macs
(or even other operating systems). For the most up-to-date information on software compatibility and relevant differences between Windows PC versions and Mac versions, please refer
to the Software Compatibility link at www.mhhe.com/Hillier6e.
We should point out that Excel is not designed for dealing with the really large management science models that occasionally arise in practice. More powerful software packages
that are not based on spreadsheets generally are used to solve such models instead. However, management science teams, not managers, primarily use these sophisticated packages
(including using modeling languages to help input the large models). Since this book is aimed
mainly at future managers rather than future management scientists, we will not have you use
these packages.
To alert you to relevant learning aids available, the end of each chapter has a list entitled
“Learning Aids for This Chapter.”
1.6 Summary
Management science is a discipline area that attempts to aid managerial decision making by applying a
scientific approach to managerial problems that involve quantitative factors. The rapid development of
this discipline began in the 1940s and 1950s. The onslaught of the computer revolution has since continued to give great impetus to its growth. Further impetus now is being provided by the widespread use
of spreadsheet software, which greatly facilitates the application of management science by managers
and others.
A major management science study involves conducting a systematic investigation that includes careful data gathering, developing and testing hypotheses about the problem (typically in the form of a mathematical model), and applying sound logic in the subsequent analysis. The management science team
then presents its recommendations to the managers who must make the decisions about how to resolve
the problem. Smaller studies might be done by managers themselves with the aid of spreadsheets.
A major part of a typical management science study involves incorporating the quantitative factors
into a mathematical model (perhaps incorporated into a spreadsheet) and then applying mathematical
20 Chapter One Introduction
procedures to solve the model. Such a model uses decision variables to represent the quantifiable decisions to be made. An objective function expresses the appropriate measure of performance in terms of
these decision variables. The constraints of the model express the restrictions on the values that can
be assigned to the decision variables. The parameters of the model are the constants that appear in the
objective function and the constraints. An example involving break-even analysis was used to illustrate
a mathematical model.
Now that we have entered the era of big data, an exciting development in recent years has been the
rise of analytics for transforming data into insight for making better decisions. The increasing momentum of the analytics movement continues to be very impressive. Analytics draws on various quantitative
decision sciences. For example, it makes heavy use of data science and forecasting, but the powerful
optimization techniques of management science also are particularly key tools. Because management
science is at the core of advanced analytics, its usage should continue to grow rapidly.
Management science has had an impressive impact on improving the efficiency of numerous organizations around the world. In fact, many award-winning applications have resulted in annual savings in
the millions, tens of millions, or even hundreds of millions of dollars.
The focus of this book is on emphasizing what an enlightened future manager needs to learn from
a management science course. Therefore, the book revolves around modeling as an aid to managerial
decision making. Many case studies (within the chapters) and cases (at the end of chapters) are used to
better convey these ideas.
Glossary
break-even point The production and sales
volume for a product that must be exceeded to
achieve a profit. (Section 1.2), 9
analytics A discipline closely related to
­management science that makes extensive use
of data to analyze trends, make forecasts, and
apply optimization techniques. (Sections 1.1
and 1.3), 3 and 12-14
Connect A web-based teaching and learning
platform that provides both SmartBook and a full
range of instructor resources. (Section 1.5), 18
constraint An inequality or equation in a mathematical model that expresses some restrictions
on the values that can be assigned to the decision
variables. (Section 1.2), 10
decision support system An interactive computerbased system that aids managerial decision making. (Section 1.1), 5
decision variable An algebraic variable that
represents a quantifiable decision to be made.
(Section 1.2), 8
mathematical model An approximate representation of, for example, a business problem that is
expressed in terms of mathematical symbols and
expressions. (Section 1.1), 4
Learning Aids for This Chapter
All learning aids are available at www.mhhe.com/Hillier6e.
Excel Files:
Special Products Co. Example
model An approximate representation of something. (Section 1.1), 4
objective function A mathematical expression
in a model that gives the measure of performance
for a problem in terms of the decision variables.
(Section 1.2), 10
operations research The traditional name for
management science that still is widely used outside of business schools. (Section 1.1), 3
parameter One of the constants in a mathematical model. (Section 1.2), 10
range name A descriptive name given to a cell
or range of cells that immediately identifies what
is there. (Section 1.2), 7
SmartBook A web-based tool within Connect
that provides personalized instruction to help students learn the material better. (Section 1.5), 18
spreadsheet model An approximate representation of, for example, a business problem that is
laid out on a spreadsheet in a way that facilitates
analysis of the problem. (Section 1.1), 4
what-if analysis Analysis of how the recommendations of a model might change if any of
the estimates providing the numbers in the model
eventually need to be corrected. (Section 1.2), 10
Chapter 1 Solved Problem 21
Solved Problem
The solution is available at www.mhhe.com/Hillier6e.
1.S1. Make or Buy?
Power Notebooks, Inc., plans to manufacture a new line of
notebook computers. Management is trying to decide whether
to purchase the LCD screens for the computers from an outside
supplier or to manufacture the screens in-house. The screens
cost $100 each from the outside supplier. To set up the assembly process required to produce the screens in-house would cost
$100,000. The company could then produce each screen for $75.
The number of notebooks that eventually will be produced (Q) is
unknown at this point.
a. Set up a spreadsheet that will display the total cost of both options for any value of Q. Use trial and error with the spreadsheet to determine the range of production volumes for which
each alternative is best.
b. Use a graphical procedure to determine the break-even point
for Q (i.e., the quantity at which both options yield the same
cost).
c. Use an algebraic procedure to determine the break-even point
for Q.
Problems
1.1. The manager of a small firm is considering whether to
produce a new product that would require leasing some special
equipment at a cost of $20,000 per month. In addition to this
leasing cost, a production cost of $10 would be incurred for each
unit of the product produced. Each unit sold would generate $20
in revenue.
Develop a mathematical expression for the monthly profit
that would be generated by this product in terms of the number
of units produced and sold per month. Then determine how large
this number needs to be each month to make it profitable to produce the product.
1.2. Management of the Toys R4U Company needs to decide
whether to introduce a certain new novelty toy for the upcoming Christmas season, after which it would be discontinued.
The total cost required to produce and market this toy would
be $500,000 plus $15 per toy produced. The company would
receive revenue of $35 for each toy sold.
a. Assuming that every unit of this toy that is produced
is sold, write an expression for the profit in terms of
the number produced and sold. Then find the breakeven point that this number must exceed to make it
worthwhile to introduce this toy.
b. Now assume that the number that can be sold might
be less than the number produced. Write an expression for the profit in terms of these two numbers.
c. Formulate a spreadsheet that will give the profit in
part b for any values of the two numbers.
d. Write a mathematical expression for the constraint
that the number produced should not exceed the
number that can be sold.
1.3. A reliable sales forecast has been obtained indicating
that the Special Products Company (see Section 1.2) would be
able to sell 30,000 iWatches, which appears to be enough to
justify introducing this new product. However, management is
concerned that this conclusion might change if more accurate
estimates were available for the research-and-development cost,
the marginal production cost, and the unit revenue. Therefore,
before a final decision is made, management wants what-if analysis done on these estimates.
Use the spreadsheet from Figure 1.3 (see this chapter’s Excel
files at www.mhhe.com/Hillier6e) and trial-and-error to perform
what-if analysis by independently investigating each of the following questions.
a. How large can the research-and-development cost
be before the watches cease to be profitable?
b. How large can the marginal production cost be
before the watches cease to be profitable?
c. How small can the unit revenue be before the
watches cease to be profitable?
1.4. Reconsider the problem facing the management of the
Special Products Company as presented in Section 1.2.
A more detailed investigation now has provided better estimates of the data for the problem. The research-and-development
cost still is estimated to be $10 million, but the new estimate of
the marginal production cost is $1,300. The revenue from each
watch sold now is estimated to be $1,700.
a. Use a graphical procedure to find the new breakeven point.
b. Use an algebraic procedure to find the new breakeven point.
c. State the mathematical model for this problem with
the new data.
d. Incorporate this mathematical model into a spreadsheet with a sales forecast of 30,000. Use this
spreadsheet model to find the new break-even point,
and then determine the production quantity and the
estimated total profit indicated by the model.
e. Suppose that management fears that the sales forecast may be overly optimistic and so does not want
to consider producing more than 20,000 watches.
Use the spreadsheet from part d to determine what
the production quantity should be and the estimated
total profit that would result.
1.5. The Best-for-Less Corp. supplies its two retail outlets
from its two plants. Plant A will be supplying 30 shipments next
month. Plant B has not yet set its production schedule for next
month but has the capacity to produce and ship any amount up
to a maximum of 50 shipments. Retail outlet 1 has submitted
its order for 40 shipments for next month. Retail outlet 2 needs
a minimum of 25 shipments next month but would be happy
to receive more. The production costs are the same at the two
22 Chapter One Introduction
plants but the shipping costs differ. The shipping cost per shipment from each plant to each retail outlet is given below, along
with a summary of the other data.
The distribution manager, Jennifer Lopez, now needs to
develop a plan for how many shipments to send from each plant
to each of the retail outlets next month. Her objective is to minimize the total shipping cost.
Unit Shipping Cost
Plant A
Plant B
Retail Outlet 1
Retail Outlet 2
Supply
$700
$800
$400
$600
= 30 shipments
≤ 50 shipments
Needed = 40 shipments
≥ 25 shipments
a. Identify the individual decisions that Jennifer needs
to make. For each of these decisions, define a decision variable to represent the decision.
b. Write a mathematical expression for the total shipping cost in terms of the decision variables.
c. Write a mathematical expression for each of the
constraints on what the values of the decision variables can be.
d. State a complete mathematical model for Jennifer’s
problem.
e. What do you think Jennifer’s shipping plan should
be? Explain your reasoning. Then express your
shipping plan in terms of the decision variables.
1.6. The Water Sports Company soon will be producing and marketing a new model line of motor boats. The production manager,
Michael Jensen, now is facing a make-or-buy decision regarding
the outboard motor to be installed on each of these boats. Based on
the total cost involved, should the motors be produced internally
or purchased from a vendor? Producing them internally would
require an investment of $1 million in new facilities as well as a
production cost of $1,600 for each motor produced. If purchased
from a vendor instead, the price would be $2,000 per motor.
Michael has obtained a preliminary forecast from the company’s marketing division that 3,000 boats in this model line will
be sold.
a. Use spreadsheets to display and analyze Michael’s
two options. Which option should be chosen?
b. Michael realizes from past experience that preliminary sales forecasts are quite unreliable, so he wants
to check on whether his decision might change if a
more careful forecast differed significantly from the
preliminary forecast. Determine a break-even point
for the production and sales volume below which
the buy option is better and above which the make
option is better.
1.7. Reconsider the Special Products Company problem presented in Section 1.2.
Although the company is well qualified to do most of the
work in producing the iWatch, it currently lacks much expertise
in one key area, namely, developing and producing a miniature
camera to be embedded into the iWatch. Therefore, management
now is considering contracting out this part of the job to another
company that has this expertise. If this were done, the Special
Products Company would reduce its research-and-development
cost to $5 million, as well as reduce its marginal production
cost to $750. However, the Special Products Company also
would pay this other company $500 for each miniature camera
and so would incur a total marginal cost of $1,250 (including
its payment to the other company) while still obtaining revenue
of $2,000 for each watch produced and sold. However, if the
company does all the production itself, all the data presented in
Section 1.2 still apply. After obtaining an analysis of the sales
potential, management believes that 30,000 watches can be sold.
Management now wants to determine whether the make
option (do all the development and production internally) or the
buy option (contract out the development and production of the
miniature cameras) is better.
a. Use a spreadsheet to display and analyze the buy
option. Show the relevant data and financial output,
including the total profit that would be obtained by
producing and selling 30,000 watches.
b. Figure 1.3 shows the analysis for the make option.
Compare these results with those from part a to
determine which option (make or buy) appears to be
better.
c. Another way to compare these two options is to
find a break-even point for the production and sales
volume, below which the buy option is better and
above which the make option is better. Begin this
process by developing an expression for the difference in profit between the make and buy options in
terms of the number of grandfather clocks to produce for sale. Thus, this expression should give the
incremental profit from choosing the make option
rather than the buy option, where this incremental
profit is 0 if 0 watches are produced but otherwise
is negative below the break-even point and positive
above the break-even point. Using this expression as
the objective function, state the overall mathematical model (including constraints) for the problem
of determining whether to choose the make option
and, if so, how many units of the LCD display (one
per watch) to produce.
d. Use a graphical procedure to find the break-even
point described in part c.
e. Use an algebraic procedure to find the break-even
point described in part c.
f. Use a spreadsheet model to find the break-even
point described in part c. What is the conclusion
about what the company should do?
1.8. Select one of the applications of management science listed
in Table 1.1. Read the article that is referenced in the application vignette presented in the section shown in the third column.
(A link to all these articles is at www.mhhe.com/­Hillier6e.)
Write a two-page summary of the application and the benefits
(including nonfinancial benefits) it provided.
1.9. Select three of the applications of management science listed
in Table 1.1. For each one, read the article that is referenced in the
application vignette presented in the section shown in the third column. (A link to all these articles is at www.mhhe.com/Hillier6e.)
For each one, write a one-page summary of the application and the
benefits (including nonfinancial benefits) it provided.
Case 1-1 Keeping Time 23
Case 1-1
Keeping Time
Founded nearly 50 years ago by Alfred Lester-Smith, ­Beautiful
Clocks specializes in developing and marketing a diverse line
of large ornamental clocks for the finest homes. Tastes have
changed over the years, but the company has prospered by continually updating its product line to satisfy its affluent clientele.
The Lester-Smith family continues to own a majority share of
the company and the grandchildren of Alfred Lester-Smith now
hold several of the top managerial positions. One of these grandchildren is Meredith Lester-Smith, the new CEO of the company.
Meredith feels a great responsibility to maintain the family
heritage with the company. She realizes that the company needs
to continue to develop and market exciting new products. Since
the 50th anniversary of the founding of the company is rapidly
approaching, she has decided to select a particularly special new
product to launch with great fanfare on this anniversary. But
what should it be? As she ponders this crucial decision, Meredith’s thoughts go back to the magnificent grandfather clock
that her grandparents had in their home many years ago. She had
admired the majesty of that clock as a child. How about launching a modern version of this clock?
This is a difficult decision. Meredith realizes that grandfather clocks now are largely out of style. However, if she is so
nostalgic about the memory of the grandfather clock in her
grandparents’ home, wouldn’t there be a considerable number
of other relatively wealthy couples with similar memories who
would welcome the prestige of adding the grandeur of a beautifully designed limited-edition grandfather clock in their home?
Maybe. This also would highlight the heritage and continuity of
the company. It all depends on whether there would be enough
sales potential to make this a profitable product.
Meredith had an excellent management science course as part
of her MBA program in college, so she realizes that break-even
analysis is needed to help make this decision. With this in mind,
she instructs several staff members to investigate this prospective
product further, including developing estimates of the related
costs and revenues as well as forecasting the potential sales.
One month later, the preliminary estimates of the relevant
financial figures come back. The cost of designing the grandfather clock and then setting up the production facilities to
produce this product would be approximately $250,000. There
would be only one production run for this limited-edition
grandfather clock. The additional cost for each clock produced
would be roughly $2,000. The marketing department estimates
that their price for selling the clocks can be successfully set at
about $4,500 apiece, but a firm forecast of how many clocks can
be sold at this price has not yet been obtained. However, it is
believed that the sales likely would reach into three digits.
Meredith wants all these numbers pinned down considerably
further. However, she feels that some analysis can be done now
to draw preliminary conclusions.
a. Assuming that all clocks produced are sold, develop a spreadsheet model for estimating the profit or loss from producing
any particular number of clocks.
b. Use this spreadsheet to find the break-even point by trial and
error.
c. Develop the corresponding mathematical expression for the
estimated profit in terms of the number of clocks produced.
d. Use a graphical procedure to find the break-even point.
e. Use the algebraic procedure to find the break-even point.
A fairly reliable forecast now has been obtained indicating
that the company would be able to sell 300 of the limited-­edition
grandfather clocks, which appears to be enough to justify introducing this new product. However, Meredith is concerned that
this conclusion might change if more accurate estimates were
available for the various costs and revenues. Therefore, she
wants what-if analysis done on these estimates. Perform whatif analysis by independently investigating each of the following
questions.
f. How large can the cost of designing this product and setting
up the production facilities be before the grandfather clocks
cease to be profitable?
g. How large can the production cost for each additional clock
be before the grandfather clocks cease to be profitable?
h. If both of the costs identified in parts f and g were 50% larger
than their initial estimates, would producing and selling the
grandfather clocks still be profitable?
i. How small can the price for selling each clock be before the
grandfather clocks cease to be profitable?
Now suppose that 300 grandfather clocks are produced but
only 200 are sold.
j. Would it still be profitable to produce and sell the grandfather
clocks under this circumstance?
Additional Case
An additional case for this chapter is also available at the University of Western Ontario Ivey School of Business Website, cases.
ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book.
Chapter Two
Linear Programming:
Basic Concepts
Learning Objectives
After completing this chapter, you should be able to
1. Explain what linear programming is.
2. Identify the three key questions to be addressed in formulating any spreadsheet model.
3. Name and identify the purpose of the four kinds of cells used in linear programming
spreadsheet models.
4. Formulate a basic linear programming model in a spreadsheet from a description of the
problem.
5. Present the algebraic form of a linear programming model from its formulation on a
spreadsheet.
6. Apply the graphical method to solve a two-variable linear programming problem.
7. Use Excel to solve a linear programming spreadsheet model.
The management of any organization regularly must make decisions about how to allocate its resources to various activities to best meet organizational objectives. Linear programming is a powerful problem-solving tool that aids management in making such
decisions. It is applicable to both profit-making and not-for-profit organizations, as well
as governmental agencies. The resources being allocated to activities can be, for example,
money, different kinds of personnel, and different kinds of machinery and equipment. In
many cases, a wide variety of resources must be allocated simultaneously. The activities
needing these resources might be various production activities (e.g., producing different
products), marketing activities (e.g., advertising in different media), financial activities
(e.g., making capital investments), or some other activities. Some problems might even involve
activities of all these types (and perhaps others), because they are competing for the same
resources.
You will see as we progress that even this description of the scope of linear programming is not sufficiently broad. Some of its applications go beyond the allocation of resources.
However, activities always are involved. Thus, a recurring theme in linear programming is the
need to find the best mix of activities—which ones to pursue and at what levels.
Like the other management science techniques, linear programming uses a mathematical
model to represent the problem being studied. The word linear in the name refers to the form
of the mathematical expressions in this model. Programming does not refer to computer programming; rather, it is essentially a synonym for planning. Thus, linear programming means
the planning of activities represented by a linear mathematical model.
Because it comprises a major part of management science, linear programming takes up
several chapters of this book. Furthermore, many of the lessons learned about how to apply
linear programming also will carry over to the application of other management science
techniques.
This chapter focuses on the basic concepts of linear programming.
24
2.1 A Case Study: The Wyndor Glass Co. Product-Mix Problem 25
2.1
A CASE STUDY: THE WYNDOR GLASS CO. PRODUCT-MIX PROBLEM
Jim Baker has had an excellent track record during his seven years as manager of new product
development for the Wyndor Glass Company. Although the company is a small one, it has
been experiencing considerable growth largely because of the innovative new products developed by Jim’s group. Wyndor’s president, John Hill, has often acknowledged publicly the key
role that Jim has played in the recent success of the company.
Therefore, John felt considerable confidence six months ago in asking Jim’s group to
develop the following new products:
• An 8-foot glass door with aluminum framing.
• A 4-foot × 6-foot double-hung, wood-framed window.
Although several other companies already had products meeting these specifications, John
felt that Jim would be able to work his usual magic in introducing exciting new features that
would establish new industry standards.
Background
The Wyndor Glass Co. produces high-quality glass products, including windows and glass
doors that feature handcrafting and the finest workmanship. Although the products are expensive, they fill a market niche by providing the highest quality available in the industry for the
most discriminating buyers. The company has three plants that simultaneously produce the
components of its products.
Plant 1 produces aluminum frames and hardware.
Plant 2 produces wood frames.
Plant 3 produces the glass and assembles the windows and doors.
Because of declining sales for certain products, top management has decided to revamp
the company’s product line. Unprofitable products are being discontinued, releasing production capacity to launch the two new products developed by Jim Baker’s group if management
approves their release.
The 8-foot glass door requires some of the production capacity in Plants 1 and 3, but not
Plant 2. The 4-foot × 6-foot double-hung window needs only Plants 2 and 3.
Management now needs to address two issues:
1. Should the company go ahead with launching these two new products?
2. If so, what should be the product mix—the number of units of each produced per week—
for the two new products?
Management’s Discussion of the Issues
Having received Jim Baker’s memorandum describing the two new products, John Hill now
has called a meeting to discuss the current issues. In addition to John and Jim, the meeting includes Bill Tasto, vice president for manufacturing, and Ann Lester, vice president for
marketing.
Let’s eavesdrop on the meeting.
John Hill (president): Bill, we will want to rev up to start production of these products as
soon as we can. About how much production output do you think we can achieve?
Bill Tasto (vice president for manufacturing): We do have a little available production
capacity, because of the products we are discontinuing, but not a lot. We should be able to
achieve a production rate of a few units per week for each of these two products.
John: Is that all?
Bill: Yes. These are complicated products requiring careful crafting. And, as I said, we
don’t have much production capacity available.
John: Ann, will we be able to sell several of each per week?
Ann Lester (vice president for marketing): Easily.
An Application Vignette
Swift & Company is a diversified protein-producing business
based in Greeley, Colorado. With annual sales of over $8 billion,
beef and related products are by far the largest portion of the
company’s business.
To improve the company’s sales and manufacturing performance, upper management concluded that it needed to
achieve three major objectives. One was to enable the company’s customer service representatives to talk to their more
than 8,000 customers with accurate information about the
availability of current and future inventory while considering
requested delivery dates and maximum product age upon
delivery. A second was to produce an efficient shift-level
schedule for each plant over a 28-day horizon. A third was to
accurately determine whether a plant can ship a requested
order-line-item quantity on the requested date and time
The issue is to find the most
profitable mix of the two
new products.
given the availability of cattle and constraints on the plant’s
capacity.
To meet these three challenges, a management science
team developed an integrated system of 45 linear programming models based on three model formulations to dynamically
schedule its beef-fabrication operations at five plants in real
time as it receives orders. The total audited benefits realized
in the first year of operation of this system were $12.74 million,
including $12 million due to optimizing the product mix. Other
benefits include a reduction in orders lost, a reduction in price
discounting, and better on-time delivery.
Source: A. Bixby, B. Downs, and M. Self, “A Scheduling and Capableto-Promise Application for Swift & Company,” Interfaces 36, no. 1
(January–February 2006), pp. 69–86. (A link to this article is available
at www.mhhe.com/Hillier6e.)
John: Good. Now there’s one more issue to resolve. With this limited production capacity,
we need to decide how to split it between the two products. Do we want to produce the same
number of both products? Or mostly one of them? Or even just produce as much as we can
of one and postpone launching the other one for a little while?
Jim Baker (manager of new product development): It would be dangerous to hold one
of the products back and give our competition a chance to scoop us.
Ann: I agree. Furthermore, launching them together has some advantages from a marketing standpoint. Since they share a lot of the same special features, we can combine the
advertising for the two products. This is going to make a big splash.
John: OK. But which mixture of the two products is going to be most profitable for the
company?
Bill: I have a suggestion.
John: What’s that?
Bill: A couple times in the past, our Management Science Group has helped us with these
same kinds of product-mix decisions, and they’ve done a good job. They ferret out all the
relevant data and then dig into some detailed analysis of the issue. I’ve found their input very
helpful. And this is right down their alley.
John: Yes, you’re right. That’s a good idea. Let’s get our Management Science Group
working on this issue. Bill, will you coordinate with them?
The meeting ends.
The Management Science Group Begins Its Work
At the outset, the Management Science Group spends considerable time with Bill Tasto to
clarify the general problem and specific issues that management wants addressed. A particular concern is to ascertain the appropriate objective for the problem from management’s
viewpoint. Bill points out that John Hill posed the issue as determining which mixture of the
two products is going to be most profitable for the company.
Therefore, with Bill’s concurrence, the group defines the key issue to be addressed as follows.
Question: Which combination of production rates (the number of units produced per week) for
the two new products would maximize the total profit from both of them?
26
The group also concludes that it should consider all possible combinations of production rates
of both new products permitted by the available production capacities in the three plants. For
example, one alternative (despite Jim Baker’s and Ann Lester’s objections) is to forgo producing one of the products for now (thereby setting its production rate equal to zero) in order to
produce as much as possible of the other product. (We must not neglect the possibility that
maximum profit from both products might be attained by producing none of one and as much
as possible of the other.)
2.2 Formulating the Wyndor Problem on a Spreadsheet 27
TABLE 2.1
Production Time Used
for Each Unit Produced
Data for the Wyndor
Glass Co. Product-Mix
Problem
Plant
Doors
Windows
Available per Week
1
2
3
1 hour
0
3 hours
0
2 hours
2 hours
4 hours
12 hours
18 hours
Unit profit
$300
$500
The Management Science Group next identifies the information it needs to gather to conduct this study:
1. Available production capacity in each of the plants.
2. How much of the production capacity in each plant would be needed by each product.
3. Profitability of each product.
Concrete data are not available for any of these quantities, so estimates have to be made.
Estimating these quantities requires enlisting the help of key personnel in other units of the
company.
Bill Tasto’s staff develops the estimates that involve production capacities. Specifically,
the staff estimates that the production facilities in Plant 1 needed for the new kind of doors
will be available approximately four hours per week. (During the rest of the time, Plant 1 will
continue with current products.) The production facilities in Plant 2 will be available for the
new kind of windows about 12 hours per week. The facilities needed for both products in
Plant 3 will be available approximately 18 hours per week.
The amount of each plant’s production capacity actually used by each product depends on
its production rate. It is estimated that each door will require one hour of production time in
Plant 1 and three hours in Plant 3. For each window, about two hours will be needed in Plant
2 and two hours in Plant 3.
By analyzing the cost data and the pricing decision, the Accounting Department estimates
the profit from the two products. The projection is that the profit per unit will be $300 for the
doors and $500 for the windows.
Table 2.1 summarizes the data now gathered.
The Management Science Group recognizes this as being a classic product-mix
­problem. Therefore, the next step is to develop a mathematical model—that is, a linear
programming model—to represent the problem so that it can be solved mathematically.
The next four sections focus on how to develop this model and then how to solve it to find
the most profitable mix between the two products, assuming the estimates in Table 2.1 are
accurate.
Review
Questions
2.2
1.
2.
3.
4.
What were the two issues addressed by management?
The Management Science Group was asked to help analyze which of these issues?
How did this group define the key issue to be addressed?
What information did the group need to gather to conduct its study?
FORMULATING THE WYNDOR PROBLEM ON A SPREADSHEET
Spreadsheets provide a powerful and intuitive tool for displaying and analyzing many management problems. We now will focus on how to do this for the Wyndor problem with the
popular spreadsheet package Microsoft Excel.1
1
Other spreadsheet packages with similar capabilities also are available, and the basic ideas presented
here are still applicable.
28 Chapter Two Linear Programming: Basic Concepts
FIGURE 2.1
The initial spreadsheet for
the Wyndor problem after
transferring the data in
Table 2.1 into data cells.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
Hours
6
Hours Used per Unit Produced
Available
7
Plant 1
1
0
4
8
Plant 2
0
2
12
9
Plant 3
3
2
18
Formulating a Spreadsheet Model for the Wyndor Problem
Excel Tip: Cell shading
and borders can be added
by using the borders button
and the fill color button
in the Font Group of the
Home tab.
Excel Tip: See the margin
notes in Section 1.2 for tips
on adding range names.
These are the three key
questions to be addressed in
formulating any spreadsheet
model.
Figure 2.1 displays the Wyndor problem by transferring the data in Table 2.1 onto a spreadsheet. (Columns E and F are being reserved for later entries described below.) We will refer
to the cells showing the data as data cells. To distinguish the data cells from other cells in
the spreadsheet, they are shaded light blue. (In the textbook figures, the light blue shading
appears as light gray.) The spreadsheet is made easier to interpret by using range names.
(As mentioned in Section 1.2, a range name is simply a descriptive name given to a cell or
range of cells that immediately identifies what is there. Excel allows you to use range names
instead of the corresponding cell addresses in Excel equations, since this usually makes the
equations much easier to interpret at a glance.) The data cells in the Wyndor Glass Co. problem are given the range names UnitProfit (C4:D4), HoursUsedPerUnitProduced (C7:D9), and
HoursAvailable (G7:G9). To enter a range name, first select the range of cells, then click
in the name box on the left of the formula bar above the spreadsheet and type a name. (See
­Appendix A for further details about defining and using range names.)
Three questions need to be answered to begin the process of using the spreadsheet to formulate a mathematical model (in this case, a linear programming model) for the problem.
1. What are the decisions to be made?
2. What are the constraints on these decisions?
3. What is the overall measure of performance for these decisions?
The preceding section described how Wyndor’s Management Science Group spent considerable time with Bill Tasto, vice president for manufacturing, to clarify management’s view of
their problem. These discussions provided the following answers to these questions.
Some students find it
helpful to organize their
thoughts by answering
these three key questions
before beginning to formulate the spreadsheet model.
The changing cells contain
the decisions to be made.
1. The decisions to be made are the production rates (number of units produced per week) for
the two new products.
2. The constraints on these decisions are that the number of hours of production time used
per week by the two products in the respective plants cannot exceed the number of hours
available.
3. The overall measure of performance for these decisions is the total profit per week from
the two products.
Figure 2.2 shows how these answers can be incorporated into the spreadsheet. Based on
the first answer, the production rates of the two products are placed in cells C12 and D12 to
locate them in the columns for these products just under the data cells. Since we don’t know
yet what these production rates should be, they are just entered as zeroes in Figure 2.2. (Actually, any trial solution can be entered, although negative production rates should be excluded
since they are impossible.) Later, these numbers will be changed while seeking the best mix
of production rates. Therefore, these cells containing the decisions to be made are called
changing cells. To highlight the changing cells, they are shaded bright yellow with a light
border. (In the textbook figures, the bright yellow appears as gray.) The changing cells are
given the range name UnitsProduced (C12:D12).
2.2 Formulating the Wyndor Problem on a Spreadsheet 29
FIGURE 2.2
The complete spreadsheet
for the Wyndor problem
with an initial trial solution (both production rates
equal to zero) entered into
the changing cells (C12
and D12).
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
0
≤
4
8
Plant 2
0
2
0
≤
12
9
Plant 3
3
2
0
≤
18
Doors
Windows
Total Profit
0
0
$0
10
11
12
The colon in C7:D9 is
Excel shorthand for the
range from C7 to D9; that
is, the entire block of cells
in column C or D and in
row 7, 8, or 9.
Units Produced
Using the answer to the second question, the total number of hours of production time used
per week by the two products in the respective plants is entered in cells E7, E8, and E9, just to
the right of the corresponding data cells. The total number of production hours depends on the
production rates of the two products, so this total is zero when the production rates are zero.
With positive production rates, the total number of production hours used per week in a plant
is the sum of the production hours used per week by the respective products. The production
hours used by a product is the number of hours needed for each unit of the product times the
number of units being produced. Therefore, when positive numbers are entered in cells C12
and D12 for the number of doors and windows to produce per week, the data in cells C7:D9
are used to calculate the total production hours per week as follows:
Production hours in Plant 1 = 1(# of doors) + 0(# of windows)
Production hours in Plant 2 = 0(# of doors) + 2(# of windows)
Production hours in Plant 3 = 3(# of doors) + 2(# of windows)
Consequently, the Excel equations for the three cells in column E are
E7 = C7*C12 + D7*D12
E8 = C8*C12 + D8*D12
E9 = C9*C12 + D9*D12
Output cells show quantities that are calculated from
the changing cells.
The SUMPRODUCT function is used extensively in
linear programming spreadsheet models.
where each asterisk denotes multiplication. Since each of these cells provides output that
depends on the changing cells (C12 and D12), they are called output cells.
Notice that each of the equations for the output cells involves the sum of two products.
There is a function in Excel called SUMPRODUCT that will sum up the product of each
of the individual terms in two different ranges of cells when the two ranges have the same
number of rows and the same number of columns. Each product being summed is the product
of a term in the first range and the term in the corresponding location in the second range.
For example, consider the two ranges, C7:D7 and C12:D12, so that each range has one row
and two columns. In this case, SUMPRODUCT (C7:D7, C12:D12) takes each of the individual terms in the range C7:D7, multiplies them by the corresponding term in the range
C12:D12, and then sums up these individual products, just as shown in the first equation
above. Applying the range name for UnitsProduced (C12:D12), the formula for E7 becomes
=SUMPRODUCT(C7:D7, UnitsProduced). Although optional with such short equations, the
SUMPRODUCT function is especially handy as a shortcut for entering longer equations.
The formulas in the output cells E7:E9 are very similar. Rather than typing each of these
formulas separately into the three cells, it is quicker (and less prone to typos) to type the
formula just once in E7 and then copy the formula down into cells E8 and E9. To do this,
first enter the formula = SUMPRODUCT(C7:D7, UnitsProduced) in cell E7. Then select cell
30 Chapter Two Linear Programming: Basic Concepts
You can make the column
absolute and the row relative (or vice versa) by putting a $ sign in front of only
the letter (or number) of the
cell reference.
One easy way to enter a ≤
(or ≥) in a spreadsheet is to
type < (or >) with underlining turned on.
Excel Tip: After entering
a cell reference, repeatedly
pressing the F4 key (or
command-T on a Mac) will
rotate among the four possibilities of relative and absolute references (e.g., C12,
$C$12, C$12, $C12).
E7 and drag the fill handle (the small box on the lower right corner of the cell cursor) down
through cells E8 and E9.
When copying formulas, it is important to understand the difference between relative and
absolute references. In the formula in cell E7, the reference to cells C7:D7 is based upon the
relative position to the cell containing the formula. In this case, the relative position is the two
cells in the same row and immediately to the left. This is known as a relative reference. When
this formula is copied to new cells using the fill handle, the reference is automatically adjusted
to refer to the new cell(s) at the same relative location (the two cells in the same row and immediately to the left). The formula in E8 becomes = SUMPRODUCT(C8:D8, UnitsProduced) and
the formula in E9 becomes = SUMPRODUCT(C9:D9, UnitsProduced). This is exactly what
we want, since we always want the hours used at a given plant to be based upon the hours used
per unit produced at that same plant (the two cells in the same row and immediately to the left).
In contrast, the reference to the UnitsProduced in E7 is called an absolute reference.
These references do not change when they are filled into other cells but instead always refer to
the same absolute cell locations.
To make a relative reference, simply enter the cell addresses (e.g., C7:D7). By contrast,
references referred to by a range name are treated as absolute references. Another way to
make an absolute reference to a range of cells is to put $ signs in front of the letter and number
of the cell reference (e.g., $C$12:$D$12). See Appendix A for more details about relative and
absolute referencing and copying formulas.
Next, ≤ signs are entered in cells F7, F8, and F9 to indicate that each total value to their left
cannot be allowed to exceed the corresponding number in column G. (On the computer ≤ (or ≥)
is often represented as < = (or > =), since there is no ≤ (or ≥) key on the keyboard.) The
spreadsheet still will allow you to enter trial solutions that violate the ≤ signs. However, these
≤ signs serve as a reminder that such trial solutions need to be rejected if no changes are made
in the numbers in column G.
Finally, since the answer to the third question is that the overall measure of performance is
the total profit from the two products, this profit (per week) is entered in cell G12. Much like
the numbers in column E, it is the sum of products. Since cells C4 and D4 give the profit from
each door and window produced, the total profit per week from these products is
​Profit = $300​​(​​# of doors​)​​​ + $500​​(​​# of windows​)​​​​
Hence, the equation for cell G12 is
​G12 = SUMPRODUCT​​(​​C4:D4, C12:D12​)​​​​
Utilizing range names of TotalProfit (G12), UnitProfit (C4:D4), and UnitsProduced (C12:D12),
this equation becomes
​TotalProfit = SUMPRODUCT​​(​​UnitProfit, UnitsProduced​)​​​​
The objective cell contains
the overall measure of performance for the decisions
in the changing cells.
This is a good example of the benefit of using range names for making the resulting equation
easier to interpret.
TotalProfit (G12) is a special kind of output cell. It is our objective to make this cell as
large as possible when making decisions regarding production rates. Therefore, TotalProfit
(G12) is referred to as the objective cell. This cell is shaded orange with a heavy border. (In
the textbook figures, the orange appears as gray and is distinguished from the changing cells
by its darker shading and heavy border.)
The bottom of Figure 2.3 summarizes all the formulas that need to be entered in the Hours
Used column and in the Total Profit cell. Also shown is a summary of the range names (in
alphabetical order) and the corresponding cell addresses.
This completes the formulation of the spreadsheet model for the Wyndor problem.
With this formulation, it becomes easy to analyze any trial solution for the production
rates. Each time production rates are entered in cells C12 and D12, Excel immediately calculates the output cells for hours used and total profit. For example, Figure 2.4 shows the
spreadsheet when the production rates are set at four doors per week and three windows
per week. Cell G12 shows that this yields a total profit of $2,700 per week. Also note that
E7 = G7, E8 < G8, and E9 = G9, so the ≤ signs in column F are all satisfied. Thus, this trial
2.2 Formulating the Wyndor Problem on a Spreadsheet 31
FIGURE 2.3
The spreadsheet model
for the Wyndor problem,
including the formulas for
the objective cell TotalProfit (G12) and the other
output cells in column E,
where the goal is to maximize the objective cell.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
Unit Profit
4
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
0
≤
4
8
Plant 2
0
2
0
≤
12
9
Plant 3
3
2
0
≤
18
Doors
Windows
Total Profit
0
0
$0
10
11
12
Units Produced
Range Name
Cell
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
TotalProfit
UnitProfit
UnitsProduced
G7:G9
E7:E9
C7:D9
G12
C4:D4
C12:D12
E
5
Hours
6
7
8
9
Used
=SUMPRODUCT(C7:D7, UnitsProduced)
=SUMPRODUCT(C8:D8, UnitsProduced)
=SUMPRODUCT(C9:D9, UnitsProduced)
G
11
12
FIGURE 2.4
The spreadsheet for the
Wyndor problem with
a new trial solution
entered into the changing cells, UnitsProduced
(C12:D12).
A
1
B
C
Total Profit
=SUMPRODUCT(UnitProfit, UnitsProduced)
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
8
Plant 2
0
9
Plant 3
3
Doors
Windows
Total Profit
4
3
$2,700
4
≤
4
2
6
≤
12
2
18
≤
18
10
11
12
Units Produced
solution is feasible. However, it would not be feasible to further increase both production
rates, since this would cause E7 > G7 and E9 > G9.
Does this trial solution provide the best mix of production rates? Not necessarily. It might
be possible to further increase the total profit by simultaneously increasing one production
rate and decreasing the other. However, it is not necessary to continue using trial and error to
explore such possibilities. We shall describe in Section 2.5 how the Excel Solver can be used
to quickly find the best (optimal) solution.
This Spreadsheet Model Is a Linear Programming Model
The spreadsheet model displayed in Figure 2.3 is an example of a linear programming model.
The reason is that it possesses all the following characteristics.
32 Chapter Two Linear Programming: Basic Concepts
Characteristics of a Linear Programming Model on a Spreadsheet
1. Decisions need to be made on the levels of a number of activities, so changing cells are
used to display these levels. (The two activities for the Wyndor problem are the production
of the two new products, so the changing cells display the number of units produced per
week for each of these products.)
2. These activity levels can have any value (including fractional values) that satisfy a number
of constraints. (The production rates for Wyndor’s new products are restricted only by the
constraints on the number of hours of production time available in the three plants.)
3. Each constraint describes a restriction on the feasible values for the levels of the activities, where a constraint commonly is displayed by having an output cell on the left, a
mathematical sign (≤, ≥, or =) in the middle, and a data cell on the right. (Wyndor’s three
constraints involving hours available in the plants are displayed in Figures 2.2–2.4 by having output cells in column E, ≤ signs in column F, and data cells in column G.)
4. The decisions on activity levels are to be based on an overall measure of performance, which
is entered in the objective cell. The goal is to either maximize the objective cell or minimize the
objective cell, depending on the nature of the measure of performance. (Wyndor’s overall measure of performance is the total profit per week from the two new products, so this measure has
been entered in the objective cell G12, where the goal is to maximize this objective cell.)
5. The Excel equation for each output cell (including the objective cell) can be expressed as
a SUMPRODUCT function,2 where each term in the sum is the product of a data cell and
a changing cell. (The bottom of Figure 2.3 shows how a SUMPRODUCT function is used
for each output cell for the Wyndor problem.)
Linear programming models are not the only models that can have characteristics 1, 3, and 4.
However, characteristics 2 and 5 are key assumptions of linear programming. Therefore, these
are the two key characteristics that together differentiate a linear programming model from
other kinds of mathematical models that can be formulated on a spreadsheet.
Characteristic 2 rules out situations where the activity levels need to have integer values.
For example, such a situation would arise in the Wyndor problem if the decisions to be made
were the total numbers of doors and windows to produce (which must be integers) rather than
the numbers per week (which can have fractional values since a door or window can be started
in one week and completed in the next week). When the activity levels do need to have integer values, a similar kind of model (called an integer programming model) is used instead by
making a small adjustment on the spreadsheet, as will be illustrated in Section 3.2.
Characteristic 5 describes the so-called proportionality assumption of linear programming,
which states that each term in an output cell must be proportional to a particular changing cell.
This prohibits those cases where the Excel equation for an output cell cannot be expressed as
a SUMPRODUCT function. To illustrate such a case, suppose that the weekly profit from
producing Wyndor’s new windows can be more than doubled by doubling the production rate
because of economies in marketing larger amounts. This would mean that this weekly profit
is not simply proportional to this production rate, so the Excel equation for the objective cell
would need to be more complicated than a SUMPRODUCT function. Consideration of how
to formulate such models will be deferred to Chapter 8.
Additional examples of formulating linear programming models on a spreadsheet are provided in Section 2.7 and in both of this chapter’s solved problems presented at the end of the
chapter. (All of these examples also will illustrate the procedure described in the next section
for using algebra to formulate linear programming models.)
Summary of the Formulation Procedure
The procedure used to formulate a linear programming model on a spreadsheet for the
Wyndor problem can be adapted to many other problems as well. Here is a summary of the
steps involved in the procedure.
1. Gather the data for the problem (such as summarized in Table 2.1 for the Wyndor problem).
2. Enter the data into data cells on a spreadsheet.
2
There also are some special situations where a SUM function can be used instead because all the
­numbers that would have gone into the corresponding data cells are 1’s.
2.3 The Mathematical Model in the Spreadsheet 33
3. Identify the decisions to be made on the levels of activities and designate changing cells
for displaying these decisions.
4. Identify the constraints on these decisions and introduce output cells as needed to specify
these constraints.
5. Choose the overall measure of performance to be entered into the objective cell.
6. Use a SUMPRODUCT function to enter the appropriate value into each output cell (including the objective cell).
This procedure does not spell out the details of how to set up the spreadsheet. There generally are alternative ways of doing this rather than a single “right” way. One of the great
strengths of spreadsheets is their flexibility for dealing with a wide variety of problems.
Review
Questions
2.3
1. What are the three questions that need to be answered to begin the process of formulating a
linear programming model on a spreadsheet?
2. What are the roles for the data cells, the changing cells, the output cells, and the objective cell
when formulating such a model?
3. What is the form of the Excel equation for each output cell (including the objective cell) when
formulating such a model?
THE MATHEMATICAL MODEL IN THE SPREADSHEET
A linear programming
model can be formulated
either as a spreadsheet
model or as an algebraic
model.
There are two widely used methods for formulating a linear programming model. One is to
formulate it directly on a spreadsheet, as described in the preceding section. The other is to
use algebra to present the model. The two versions of the model are equivalent. The only difference is whether the language of spreadsheets or the language of algebra is used to describe
the model. Both versions have their advantages, and it can be helpful to be bilingual. For
example, the two versions lead to different, but complementary, ways of analyzing problems
like the Wyndor problem (as discussed in the next two sections). Since this book emphasizes
the spreadsheet approach, we will only briefly describe the algebraic approach.
Formulating the Wyndor Model Algebraically
The reasoning for the algebraic approach is similar to that for the spreadsheet approach. In
fact, except for making entries on a spreadsheet, the initial steps are just as described in the
preceding section for the Wyndor problem.
1. Gather the relevant data (Table 2.1 in Section 2.1).
2. Identify the decisions to be made (the production rates for the two new products).
3. Identify the constraints on these decisions (the production time used in the respective
plants cannot exceed the amount available).
4. Identify the overall measure of performance for these decisions (the total profit from the
two products).
5. Convert the verbal description of the constraints and measure of performance into quantitative expressions in terms of the data and decisions (see below).
To start performing step 5, note that Table 2.1 indicates that the number of hours of production time available per week for the two new products in the respective plants are 4, 12,
and 18. Using the data in this table for the number of hours used per door or window produced
then leads to the following quantitative expressions for the constraints:
Plant 1:       (# of doors)
≤ 4
Plant
​​
2:​ ​
​  2(# of windows)​  ​  ≤​ ​  12​​​​
Plant 3:       3​​(​​# of doors​)​​​ + 2​​(​​# of windows​)​​​ ≤ 18
In addition, negative production rates are impossible, so two other constraints on the decisions are
​​(​​# of doors​)​​​ ≥ 0​  ​ ​(​​# of windows​)​​​ ≥ 0​​
The overall measure of performance has been identified as the total profit from the two
products. Since Table 2.1 gives the unit profits for doors and windows as $300 and $500,
34 Chapter Two Linear Programming: Basic Concepts
respectively, the expression obtained in the preceding section for the total profit per week
from these products is
​Profit = $300​​(​​# of doors​)​​​ + $500​​(​​# of windows​)​​​​
The goal is to make the decisions (number of doors and number of windows) so as to maximize this profit, subject to satisfying all the constraints identified above.
To state this objective in a compact algebraic model, we introduce algebraic symbols to
represent the measure of performance and the decisions. Let
P = Profit (total profit per week from the two products, in dollars)
D = # of doors (number of the special new doors to be produced per week)
W = # of windows (number of the special new windows to be produced per week)
Substituting these symbols into the above expressions for the constraints and the measure of
performance (and dropping the dollar signs in the latter expression), the linear programming
model for the Wyndor problem now can be written in algebraic form as shown below.
Algebraic Model
Choose the values of D and W so as to maximize
​P = 300D + 500W​
subject to satisfying all the following constraints:
D
≤ 4
​​
2W​ ​  ≤​ ​  12​​​​
3D + 2W ≤ 18
and
​​D ≥ 0​
W ≥ 0​​
Terminology for Linear Programming Models
Much of the terminology of algebraic models also is sometimes used with spreadsheet models. Here are the key terms for both kinds of models in the context of the Wyndor problem.
1. D and W (or C12 and D12 in Figure 2.3) are the decision variables.
2. 300D + 500W [or SUMPRODUCT (UnitProfit, UnitsProduced)] is the objective
function.
3. P (or G12) is the value of the objective function (or objective value for short).
4. D ≥ 0 and W ≥ 0 (or C12 ≥ 0 and D12 ≥ 0) are called the nonnegativity constraints (or
nonnegativity conditions).
5. The other constraints are referred to as functional constraints (or structural constraints).
6. The parameters of the model are the constants in the algebraic model (the numbers in the
data cells).
7. Any choice of values for the decision variables (regardless of how desirable or undesirable
the choice) is called a solution for the model.
8. A feasible solution is one that satisfies all the constraints, whereas an infeasible
­solution violates at least one constraint.
9. The best feasible solution, the one that maximizes P (or G12), is called the optimal
­solution. (It is possible to have a tie for the best feasible solution, in which case all the tied
solutions are called optimal solutions.)
Comparisons
So what are the relative advantages of algebraic models and spreadsheet models? An algebraic
model provides a very concise and explicit statement of the problem. Sophisticated software
packages that can solve huge problems generally are based on algebraic models because of
both their compactness and their ease of use in rescaling the size of a problem. Management
2.4 The Graphical Method for Solving Two-Variable Problems 35
Management scientists
often use algebraic models,
but managers generally prefer spreadsheet models.
science practitioners with an extensive mathematical background find algebraic models very
useful. For others, however, spreadsheet models are far more intuitive. Both managers and
business students training to be managers generally live with spreadsheets, not algebraic models. Therefore, the emphasis throughout this book is on spreadsheet models.
Review
Questions
1. When formulating a linear programming model, what are the initial steps that are the same
with either a spreadsheet formulation or an algebraic formulation?
2. When formulating a linear programming model algebraically, algebraic symbols need to be
introduced to represent which kinds of quantities in the model?
3. What are decision variables for a linear programming model? The objective function? Nonnegativity constraints? Functional constraints?
4. What is meant by a feasible solution for the model? An optimal solution?
2.4 THE GRAPHICAL METHOD FOR SOLVING TWO-VARIABLE PROBLEMS
graphical method The
graphical method provides
helpful intuition about linear programming.
Linear programming problems having only two decision variables, like the Wyndor problem,
can be solved by a graphical method.
Although this method cannot be used to solve problems with more than two decision variables (and most linear programming problems have far more than two), it still is well worth
learning. The procedure provides geometric intuition about what linear programming is and
what it is trying to achieve. This intuition is very helpful in analyzing linear programming
models, including larger problems that cannot be solved directly by the graphical method. For
example, it enables visualizing what is happening when performing what-if analysis for linear
programming (the subject of Chapter 5).
It is more convenient to apply the graphical method to the algebraic version of the linear programming model rather than the spreadsheet version. We shall briefly illustrate the
method by using the algebraic model obtained for the Wyndor problem in the preceding
section. (A far more detailed description of the graphical method, including its application
to the Wyndor problem, is provided in the supplement to this chapter that is available at
www.mhhe.com/Hillier6e.) For this purpose, keep in mind that
D = Production rate for the special new doors (the number in changing cell C12 of the
spreadsheet)
W = Production rate for the special new windows (the number in changing cell D12 of
the spreadsheet)
The key to the graphical method is the fact that possible solutions can be displayed as
points on a two-dimensional graph that has a horizontal axis giving the value of D and a vertical axis giving the value of W. Figure 2.5 shows some sample points.
Notation: Either (D, W) = (2, 3) or just (2, 3) refers to the solution where D = 2 and W = 3,
as well as to the corresponding point in the graph. Similarly, (D, W) = (4, 6) means D = 4 and
W = 6, whereas the origin (0, 0) means D = 0 and W = 0.
feasible region The points
in the feasible region are
those that satisfy every
constraint.
To find the optimal solution (the best feasible solution), we first need to display graphically where the feasible solutions are. To do this, we must consider each constraint, identify
the solutions graphically that are permitted by that constraint, and then combine this information to identify the solutions permitted by all the constraints. The solutions permitted by all
the constraints are the feasible solutions and the portion of the two-dimensional graph where
the feasible solutions lie is referred to as the feasible region.
The shaded region in Figure 2.6 shows the feasible region for the Wyndor problem. We
now will outline how this feasible region was identified by considering the five constraints
one at a time.
To begin, the constraint D ≥ 0 implies that consideration must be limited to points that lie
on or to the right of the W axis. Similarly, the constraint W ≥ 0 restricts consideration to the
points on or above the D axis.
36 Chapter Two Linear Programming: Basic Concepts
FIGURE 2.5
W
Graph showing the points
(D, W) = (2, 3) and
(D, W) = (4, 6) for the
Wyndor Glass Co.
product-mix problem.
Production Rate (units per week) for Windows
8
–1
(4, 6)
6
5
4
A product mix of
D = 2 and W = 3
(2, 3)
3
2
1
–2
A product mix of
D = 4 and W = 6
7
Origin
0
1
–1
2
3
4
5
6
7
Production Rate (units per week) for Doors
8
D
–2
Next, consider the first functional constraint, D ≤ 4, which limits the usage of Plant 1 for
producing the special new doors to a maximum of four hours per week. The solutions permitted by this constraint are those that lie on, or to the left of, the vertical line that intercepts the
D axis at D = 4, as indicated by the arrows pointing to the left from this line in Figure 2.6.
FIGURE 2.6
W
10
Production Rate for Windows
Graph showing how the
feasible region is formed
by the constraint boundary lines, where the
arrows indicate which
side of each line is permitted by the corresponding
constraint.
8
3D + 2W = 18
D=4
2W = 12
6
4
Feasible
region
2
0
2
4
6
Production Rate for Doors
8
D
2.4 The Graphical Method for Solving Two-Variable Problems 37
For any constraint with an
inequality sign, its constraint boundary equation
is obtained by replacing the
inequality sign by an equality sign.
The second functional constraint, 2W ≤ 12, has a similar effect, except now the boundary of its permissible region is given by a horizontal line with the equation, 2W = 12
(or W = 6), as indicated by the arrows pointing downward from this line in Figure 2.6. The
line forming the boundary of what is permitted by a constraint is sometimes referred to
as a constraint boundary line, and its equation may be called a constraint boundary
equation. Frequently, a constraint boundary line is identified by its equation.
For each of the first two functional constraints, D ≤ 4 and 2W ≤ 12, note that the equation
for the constraint boundary line (D = 4 and 2W = 12, respectively) is obtained by replacing
the inequality sign with an equality sign. For any constraint with an inequality sign (whether
a functional constraint or a nonnegativity constraint), the general rule for obtaining its constraint boundary equation is to substitute an equality sign for the inequality sign.
We now need to consider one more functional constraint, 3D + 2W ≤ 18. Its constraint
boundary equation
​3D + 2W = 18​
includes both variables, so the boundary line it represents is neither a vertical line nor a horizontal line. Therefore, the boundary line must intercept (cross through) both axes somewhere.
But where?
When a constraint boundary line is neither a vertical line nor a horizontal line, the line intercepts the D axis at the point on the line where W = 0. Similarly, the line intercepts the W axis at
the point on the line where D = 0.
Hence, the constraint boundary line 3D + 2W = 18 intercepts the D axis at the point where
W = 0.
When W = 0, 3D + 2W = 18
becomes 3D = 18
​​
​
​
​  ​  ​​​
so the intercept with the D axis is at
D= 6
Similarly, the line intercepts the W axis where D = 0.
When D = 0, 3D + 2W = 18
becomes 2W = 18
​​
​
​
​  ​  ​​​
so the intercept with the D axis is at
W= 9
The location of a slanting
constraint boundary line is
found by identifying where
it intercepts each of the two
axes.
Checking whether (0, 0)
satisfies a constraint indicates which side of the
constraint boundary line
satisfies the constraint.
Consequently, the constraint boundary line is the line that passes through these two intercept
points, as shown in Figure 2.6.
As indicated by the arrows emanating from this line in Figure 2.6, the solutions permitted
by the constraint 3D + 2W ≤ 18 are those that lie on the origin side of the constraint boundary
line 3D + 2W = 18. The easiest way to verify this is to check whether the origin itself, (D, W) =
(0, 0), satisfies the constraint.3 If it does, then the permissible region lies on the side of the
constraint boundary line where the origin is. Otherwise, it lies on the other side. In this case,
​3​​(​​0​)​​​ + 2​​(​​0​)​​​ = 0​
so (D, W) = (0, 0) satisfies
​3D + 2W ≤ 18​
(In fact, the origin satisfies any constraint with a ≤ sign and a positive right-hand side.)
A feasible solution for a linear programming problem must satisfy all the constraints
simultaneously. The arrows in Figure 2.6 indicate that the nonnegative solutions permitted by
each of these constraints lie on the side of the constraint boundary line where the origin is (or
actually on the line itself). Therefore, the feasible solutions are the nonnegative solutions that
lie nearer to the origin than all three constraint boundary lines (or actually on the line nearest
the origin at that point).
Having identified the feasible region, the final step is to find which of these feasible solutions is the best one—the optimal solution. For the Wyndor problem, the objective happens to
3
The one case where using the origin to help determine the permissible region does not work is if the
constraint boundary line passes through the origin. In this case, any other point not lying on this line can be
used just like the origin.
38 Chapter Two Linear Programming: Basic Concepts
be to maximize the total profit per week from the two products (denoted by P). Therefore, we
want to find the feasible solution (D, W) that makes the value of the objective function
​P = 300D + 500W​
as large as possible.
To accomplish this, we need to be able to locate all the points (D, W) on the graph that give
a specified value of the objective function. For example, consider a value of P = 1,500 for the
objective function. Which points (D, W) give 300D + 500W = 1,500?
This equation is the equation of a line. Just as when plotting constraint boundary lines, the
location of this line is found by identifying its intercepts with the two axes. When W = 0, this
equation yields D = 5, and similarly, W = 3 when D = 0, so these are the two intercepts, as
shown by the bottom slanting line passing through the feasible region in Figure 2.7.
P = 1,500 is just one sample value of the objective function. For any other specified value of
P, the points (D, W) that give this value of P also lie on a line called an objective function line.
An objective function line is a line whose points all have the same value of the objective
function.
For the bottom objective function line in Figure 2.7, the points on this line that lie in the
feasible region provide alternate ways of achieving an objective function value of P = 1,500.
Can we do better? Let us try doubling the value of P to P = 3,000. The corresponding objective function line
​300D + 500W = 3,000​
is shown as the middle line in Figure 2.7. (Ignore the top line for the moment.) Once again,
this line includes points in the feasible region, so P = 3,000 is achievable.
Let us pause to note two interesting features of these objective function lines for P = 1,500
and P = 3,000. First, these lines are parallel. Second, doubling the value of P from 1,500
to 3,000 also doubles the value of W at which the line intercepts the W axis from W = 3 to
W = 6. These features are no coincidence, as indicated by the following properties.
Key Properties of Objective Function Lines: All objective function lines for the same problem are parallel. Furthermore, the value of W at which an objective function line intercepts the
W axis is proportional to the value of P.
FIGURE 2.7
Graph showing three
objective function lines
for the Wyndor Glass Co.
product-mix problem,
where the top one passes
through the optimal
solution.
Production Rate W
for Windows
8
P = 3,600 = 300D + 500W
Optimal solution
P = 3,000 = 300D + 500W
6
4
P = 1,500 = 300D + 500W
(2, 6)
Feasible
region
2
0
2
4
6
Production Rate for Doors
8
10 D
2.5 Using Excel’s Solver to Solve Linear Programming Problems 39
These key properties of objective function lines suggest the strategy to follow to find the
optimal solution. We already have tried P = 1,500 and P = 3,000 in Figure 2.7 and found that
their objective function lines include points in the feasible region. Increasing P again will
generate another parallel objective function line farther from the origin. The objective function line of special interest is the one farthest from the origin that still includes a point in the
feasible region. This is the third objective function line in Figure 2.7. The point on this line
that is in the feasible region, (D, W) = (2, 6), is the optimal solution since no other feasible
solution has a larger value of P.
Optimal Solution
D = 2 (Produce 2 special new doors per week)
W = 6 (Produce 6 special new windows per week)
These values of D and W can be substituted into the objective function to find the value of P.
​P = 300D + 500W = 300​​(​​2​)​​​ + 500​​(​​6​)​​​ = 3,600​
This has been a fairly quick description of the graphical method. You can go to the supplement to this chapter that is available at www.mhhe.com/Hillier6e if you would like to see a
fuller description of how to use the graphical method to solve the Wyndor problem.
The end of Section 2.7 will provide another example of the graphical method, where the
objective in that case is to minimize instead of maximize the objective function. An additional
example also is included in the first of this chapter’s solved problems (Problem 2.S-1) presented at the end of the chapter.
Summary of the Graphical Method
The graphical method can be used to solve any linear programming problem having only two
decision variables. The method uses the following steps:
1. Draw the constraint boundary line for each functional constraint. Use the origin (or any
point not on the line) to determine which side of the line is permitted by the constraint.
2. Find the feasible region by determining where all constraints are satisfied simultaneously.
3. Determine the slope of one objective function line. All other objective function lines will
have the same slope.
4. Move a straight edge with this slope through the feasible region in the direction of improving values of the objective function. Stop at the last instant that the straight edge still passes
through a point in the feasible region. This line given by the straight edge is the optimal
objective function line.
5. A feasible point on the optimal objective function line is an optimal solution.
Review
Questions
1. The graphical method can be used to solve linear programming problems with how many
decision variables?
2. What do the axes represent when applying the graphical method to the Wyndor problem?
3. What is a constraint boundary line? A constraint boundary equation?
4. What is the easiest way of determining which side of a constraint boundary line is permitted by
the constraint?
2.5 USING EXCEL’S SOLVER TO SOLVE LINEAR PROGRAMMING PROBLEMS
The graphical method is very useful for gaining geometric intuition about linear programming, but its practical use is severely limited by only being able to solve tiny problems with
two decision variables. Another procedure that will solve linear programming problems of
any reasonable size is needed. Fortunately, Excel includes a tool called Solver that will do
this once the spreadsheet model has been formulated as described in Section 2.2. (Section 2.6
will show how the Analytic Solver software package, which includes a more advanced version
of Solver, can be used to solve this same problem.) To access Solver the first time, you need to
40 Chapter Two Linear Programming: Basic Concepts
Excel Tip: If you select cells
by clicking on them, they
will first appear in the dialog
box with their cell addresses
and with dollar signs (e.g.,
$C$9:$D$9). You can ignore
the dollar signs. Solver eventually will replace both the
cell addresses and the dollar
signs with the corresponding
range name (if a range name
has been defined for the given
cell addresses), but only after
either adding a constraint or
closing and reopening the
Solver dialog box.
Solver Tip: To select
changing cells, click and
drag across the range of
cells. If the changing cells
are not contiguous, you
can type a comma and then
select another range of
cells. Up to 200 changing
cells can be selected with
the basic version of Solver
that comes with Excel.
FIGURE 2.8
The incomplete Solver
dialog box after specifying the first components of the model for
the ­Wyndor problem.
TotalProfit (G12) is
being maximized by
changing UnitsProduced
(C12:D12). Figure 2.9
will demonstrate the
addition of constraints
and then Figure 2.10 will
demonstrate the changes
needed to specify that the
problem being considered
is a linear programming
problem. (The default
Solving Method shown
here—GRG Nonlinear—
is not applicable to linear
programming problems.)
install it. In Windows versions of Excel, choose Excel Options from the File menu, then click
on Add-Ins on the left side of the window, select Manage Excel Add-Ins at the bottom of the
window, and then press the Go button. Make sure Solver is selected in the Add-Ins dialog box,
and then it should appear on the Data tab. For Mac versions of Excel, select Add-Ins from the
Tools menu, make sure Solver is selected, then click OK; Solver should now appear on the
Data tab.
Figure 2.3 in Section 2.2 shows the spreadsheet model for the Wyndor problem. The values
of the decision variables (the production rates for the two products) are in the changing cells,
UnitsProduced (C12:D12), and the value of the objective function (the total profit per week
from the two products) is in the objective cell TotalProfit (G12). To get started, an arbitrary
trial solution has been entered by placing zeroes in the changing cells. Solver will then change
these to the optimal values after solving the problem.
This procedure is started by choosing Solver on the Data tab. Figure 2.8 shows the Solver
dialog box that is used to tell Solver where each component of the model is located on the
spreadsheet.
You have the choice of typing the range names, typing the cell addresses, or probably the
easiest method, clicking on the cells in the spreadsheet. Figure 2.8 shows the results of the last
method—clicking on the cell for TotalProfit (G12) for the objective cell. Since the goal is to
maximize the objective cell, Max also has been selected. The next entry in the Solver dialog
box identifies the changing cells, specified by selecting the cells for UnitsProduced (C12:D12).
Next, the cells containing the functional constraints need to be specified. This is done
by clicking on the Add button on the Solver dialog box. This brings up the Add Constraint
dialog box shown in Figure 2.9. The ≤ signs in cells F7, F8, and F9 of Figure 2.3 are a
reminder that the cells in HoursUsed (E7:E9) all need to be less than or equal to the corresponding cells in HoursAvailable (G7:G9). These constraints are specified for Solver by
2.5 Using Excel’s Solver to Solve Linear Programming Problems 41
The Add Constraint dialog
box is used to specify all
the functional constraints.
When solving a linear programming problem, be sure
to specify that nonnegativity constraints are needed
and that the model is linear
by choosing Simplex LP.
FIGURE 2.9
The Add Constraint dialog
box after specifying that
cells E7, E8, and E9 in
Figure 2.3 are required
to be less than or equal
to cells G7, G8, and G9,
respectively.
FIGURE 2.10
The completed Solver dialog box after specifying
the entire model in terms
of the spreadsheet.
entering HoursUsed (or E7:E9) on the left-hand side of the Add Constraint dialog box and
then entering HoursAvailable (or G7:G9) on the right-hand side. For the sign between these
two sides, there is a menu to choose between < =, =, or > =, so < = has been chosen. This
choice is needed even though ≤ signs were previously entered in column F of the spreadsheet because the Solver only uses the constraints that are specified with the Add Constraint
dialog box.
If there were more functional constraints to add, you would click on Add to bring up a new
Add Constraint dialog box. However, since there are no more in this example, the next step is
to click on OK to go back to the Solver dialog box.
Before asking Solver to solve the model, two more steps need to be taken. We need to
tell Solver that nonnegativity constraints are needed for the changing cells to reject negative production rates. We also need to specify that this is a linear programming problem
so the simplex method (the standard method used by Solver to solve linear programming
problems) can be used. This is demonstrated in Figure 2.10, where the Make Unconstrained
42 Chapter Two Linear Programming: Basic Concepts
When you define range
names for cells in your
spreadsheet, these range
names are displayed by
Solver. This makes the
model in Solver much
easier to interpret.
Solver Tip: The message
“Solver could not find a
feasible solution” means
that there are no solutions that satisfy all the
constraints. The message
“The Objective Cell values
do not converge” means
that Solver could not find
a best solution, because
better solutions always are
available (e.g., if the constraints do not prevent infinite profit). The message
“The linearity conditions
required by this LP Solver
are not satisfied” means
Simplex LP was chosen as
the Solving Method, but the
model is not linear.
FIGURE 2.11
The Solver Results dialog
box that indicates that an
optimal solution has been
found.
Variables Non-Negative option has been checked and the Solving Method chosen is
Simplex LP (rather than GRG Nonlinear or Evolutionary, which are used for solving nonlinear problems). The Solver dialog box shown in this figure now summarizes the complete model. Note that after returning from the Add Constraint dialog box, Solver has now
converted all of the cell addresses to the corresponding range names that were defined for
these cells (TotalProfit for G12, UnitsProduced for C12:D12, HoursUsed for E7:E9, and
HoursAvailable for G7:G9). Defining range names makes the model in Solver much easier
to interpret.
Now you are ready to click on Solve in the Solver dialog box, which will start the solving
of the problem in the background. After a few seconds (for a small problem), Solver will then
indicate the results. Typically, it will indicate that it has found an optimal solution, as specified in the Solver Results dialog box shown in Figure 2.11. If the model has no feasible solutions or no optimal solution, the dialog box will indicate that instead by stating that “Solver
could not find a feasible solution” or that “The Objective Cell values do not converge.”
(Section 14.1 will describe how these possibilities can occur.) The dialog box also presents
the option of generating various reports. One of these (the Sensitivity Report) will be discussed in detail in Chapter 5.
After solving the model and clicking OK in the Solver Results dialog box, Solver replaces
the original numbers in the changing cells with the optimal numbers, as shown in Figure 2.12.
Thus, the optimal solution is to produce two doors per week and six windows per week, just
as was found by the graphical method in the preceding section. The spreadsheet also indicates
the corresponding number in the objective cell (a total profit of $3,600 per week), as well as
the numbers in the output cells HoursUsed (E7:E9).
The entries needed for the Solver dialog box are summarized in the Solver Parameters
box shown on the bottom left of Figure 2.12. This more compact summary of the Solver
Parameters will be shown for all of the many models that involve the Solver throughout
the book.
At this point, you might want to check what would happen to the optimal solution if any of
the numbers in the data cells were to be changed to other possible values. This is easy to do
because Solver saves all the addresses for the objective cell, changing cells, constraints, and so
on when you save the file. All you need to do is make the changes you want in the data cells
and then click on Solve in the Solver dialog box again. (Chapter 5 will focus on this kind of
what-if analysis, including how to use the Solver’s Sensitivity Report to expedite the analysis.)
To assist you with experimenting with these kinds of changes, www.mhhe.com/Hillier6e
includes Excel files for this chapter (as for others) that provide a complete formulation and
2.6 Analytic Solver 43
FIGURE 2.12
The spreadsheet obtained
after solving the Wyndor
problem.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Doors
Windows
$300
$500
Unit Profit
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
2
≤
4
8
Plant 2
0
2
12
≤
12
9
Plant 3
3
2
18
≤
18
Doors
Windows
Total Profit
2
6
$3,600
10
11
12
Units Produced
Solver Parameters
Set Objective Cell: Total Profit
To: Max
By Changing Variable Cells:
UnitsProduced
Subject to the Constraints:
HoursUsed <= HoursAvailable
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
E
5
Hours
6
7
8
9
Used
=SUMPRODUCT(C7:D7, UnitsProduced)
=SUMPRODUCT(C8:D8, UnitsProduced)
=SUMPRODUCT(C9:D9, UnitsProduced)
G
11
12
Total Profit
=SUMPRODUCT(UnitProfit, UnitsProduced)
Range Name
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
TotalProfit
UnitProfit
UnitsProduced
Cell
G7:G9
E7:E9
C7:D9
G12
C4:D4
C12:D12
solution of the examples here (the Wyndor problem and the one in Section 2.7) in a spreadsheet format. We encourage you to “play” with these examples to see what happens with different data, different solutions, and so forth. You might also find these spreadsheets useful as
templates for homework problems.
Review
Questions
2.6
1. Which dialog box is used to enter the addresses for the objective cell and the changing cells?
2. Which dialog box is used to specify the functional constraints for the model?
3. Which options normally need to be chosen to solve a linear programming model?
ANALYTIC SOLVER
Frontline Systems, the original developer of the standard Solver included with Excel (hereafter referred to as Excel’s Solver), also has developed Premium versions of Solver that provide
greatly enhanced functionality. The company now features a particularly powerful Premium
Solver called Analytic Solver. We are excited to provide access to the Excel add-in, Analytic
Solver for Education, which includes all the features of the full-fledged Analytic Solver (also
called Analytic Solver Comprehensive) except that it is limited to considerably smaller problems than the huge problems that the full-fledged version can solve. However, since Analytic
Solver for Education can handle much larger problems than any problems you will encounter
in this book, we will simply refer to it hereafter as Analytic Solver.
44 Chapter Two Linear Programming: Basic Concepts
Instructions for obtaining a low-priced student license to download this software are
provided in both the preface and www.mhhe.com/Hillier6e. A cloud-based version of this
software is also available at AnalyticSolver.com. The cloud-based version works using a
browser, but is designed to look and feel as much as possible like the downloaded Analytic
Solver add-in for Excel.
While Excel’s Solver is sufficient for most of the problems considered in this book,
Analytic Solver includes a number of important features not available with Excel’s Solver.
Where either Excel’s Solver or Analytic Solver can be used, the book will often use the
term Solver generically to mean either Excel’s Solver or Analytic Solver. Where there are
differences, the book will include instructions for both Excel’s Solver and Analytic Solver.
The enhanced features of Analytic Solver will be highlighted as they come up throughout
the book. However, if you and your instructor prefer to focus on only using Excel’s Solver,
you will find that there is plenty of material to cover in the book that does not require the
use of Analytic Solver. In this case, you can skip this section and continue on without loss
of continuity.
When Analytic Solver is installed, two new tabs are available on the Excel ribbon named
Analytic Solver and Data Mining (along with an optional third tab named Solver Home).
Choosing the Analytic Solver tab will reveal the ribbon shown in Figure 2.13. (The Data
­Mining tab will not be used in this chapter, but will be introduced in Chapter 10.) The buttons
on this ribbon will be used to interact with Analytic Solver. This same figure also reveals a
nice feature of Analytic Solver—the Solver Options and Model Specifications pane (showing
the objective cell, changing cells, constraints, etc.)—that can be seen alongside your main
spreadsheet, with both visible simultaneously. This pane can be toggled on (to see the model)
or off (to hide the model and leave more room for the spreadsheet) by clicking on the Model
button on the far left of the Analytic Solver ribbon. Also, since the model was already set up
with Excel’s Solver in Section 2.5, it is already set up in the Model pane, with the objective
specified as TotalProfit (G12) with changing cells UnitsProduced (C12:D12) and the constraints HoursUsed (E7:E9) <= HoursAvailable (G7:G9). The data for Excel’s Solver and
Analytic Solver are compatible with each other. Making a change with one makes the same
change in the other. Thus, you can work with either Excel’s Solver or Analytic Solver, and
then go back and forth, without losing any Solver data.
If the model had not been previously set up with Excel’s Solver, the steps for doing so with
Analytic Solver are analogous to the steps used with Excel’s Solver as covered in Section 2.5.
In both cases, we need to specify the location of the objective cell, the changing cells, and
the functional constraints, and then click to solve the model. However, the user interface is
somewhat different. Analytic Solver uses the buttons on the Analytic Solver ribbon instead of
the Solver dialog box. We will now walk you through the steps to set up the Wyndor problem
in Analytic Solver.
To specify TotalProfit (G12) as the objective cell, select the cell in the spreadsheet and
then click on the Objective button on the Analytic Solver ribbon. As shown in Figure 2.14,
FIGURE 2.13
The screenshot for the
Wyndor problem that
shows both the ribbon and
the Solver Options and
Model Specifications pane
that are revealed after
choosing the tab called
Analytic Solver on the
Excel ribbon.
2.6 Analytic Solver 45
FIGURE 2.14
The screenshot for the
Wyndor problem that
shows the drop-down
menu generated by clicking on the Objective button on the Analytic Solver
ribbon after choosing
TotalProfit (G12) as the
objective cell.
Analytic Solver Tip:
Another way to add an
objective, changing cells,
or constraints using Analytic Solver is to click on
the big green plus (+) on
the Mode pane and choose
Add Objective, Add Variable, or Add Constraint,
respectively.
FIGURE 2.15
The screenshot for the
Wyndor problem that
shows the drop-down
menu generated by clicking on the Decision button
on the Analytic Solver
ribbon after choosing
UnitsProduced (C12:D12)
as the changing variable
cells.
FIGURE 2.16
The screenshot of the
Wyndor problem that
shows the drop-down
menu generated by clicking on the Constraints
button on the Analytic
Solver ribbon after choosing HoursUsed (E7:E9) as
the left-hand side of the
functional constraints.
this will drop down a menu where you can choose to minimize (Min) or maximize (Max)
the objective cell. Within the options of Min or Max are further options (Normal, Expected,
Chance). For now, we will always choose the Normal option.
To specify UnitsProduced (C12:D12) as the changing cells, select these cells in the spreadsheet and then click on the Decisions button on the Analytic Solver ribbon. As shown in
Figure 2.15, this will drop down a menu where you can choose various options (Plot, Normal,
Recourse). For linear programming, we will always choose the Normal option.
Next the functional constraints need to be specified. For the Wyndor problem, the functional
constraints are HoursUsed (E7:E9) <= HoursAvailable (G7:G9). To enter these constraints in
Analytic Solver, select the cells representing the left-hand side of these constraints (HoursUsed,
or E7:E9) and click the Constraints button on the Analytic Solver ribbon. As shown in Figure
2.16, this drops down a menu for various kinds of constraints. For linear programming functional
46 Chapter Two Linear Programming: Basic Concepts
FIGURE 2.17
The Add Constraint dialog
box that is brought up
after choosing <= in Figure 2.16. This dialog box
also shows HoursAvailable (G7:G9) as the righthand side of the functional
constraints after clicking
in the Constraint box and
choosing these cells on
the spreadsheet.
Analytic Solver Tip:
Double-clicking on any
element of the model (e.g.,
the objective, any of the
changing cells, or any of the
constraints) will bring up a
dialog box allowing you to
make changes to that part of
the model.
FIGURE 2.18
This figure shows the
result of clicking on the
HoursUsed <= Hours
Available constraint in the
Model pane prior to considering possible changes
in the constraint.
constraints, choose Normal Constraint and then the type of constraint desired (either <=, =, or
>=). For the Wyndor problem, choosing <= would then bring up the Add Constraint dialog
box shown in Figure 2.17. This is much like the Add Constraint dialog box for Excel’s Solver
(see ­Figure 2.9). HoursUsed and <= already are filled in when the Add Constraint dialog box is
brought up (because the HoursUsed cells were selected and <= was chosen under the Constraints
button menu). The Add Constraint dialog box then can be used to fill in the remaining right-hand
side of the constraint—HoursAvailable (G7:G9)—by clicking in the box labeled Constraint and
choosing these cells on the spreadsheet. Figure 2.17 shows the dialog box after this has been done.
Changes to the model can easily be made within the Model pane shown in Figure 2.13. For
example, to delete an element of the model (e.g., the objective, changing cells, or constraints),
select that part of the model and then click on the red X near the top of the Model pane. To
change an element of the model, click on that element in the Model pane. The bottom of the
Model pane will then show information about that element. For example, clicking on the Hours
Used <= HoursAvailable constraint in the Model pane will then show the information seen in
Figure 2.18. Clicking on any piece of the information will allow you to change it (e.g., you can
change <= to >=, or you can change the cell references for either side of the constraint).
Selecting the Engine tab at the top of the Model pane will show information about the algorithm that will be used to solve the problem as well as a variety of options for that algorithm.
The drop-down menu at the top will allow you to choose the algorithm. For a linear programming model (such as the Wyndor problem), you will want to choose the Standard LP/Quadratic
Engine. This is equivalent to the Simplex LP option in Excel’s Solver. To make unconstrained
variables nonnegative (as we did in Figure 2.10 with Excel’s Solver), be sure that the Assume
Nonnegative option is set to true. Figure 2.19 shows the model pane after making these selections.
2.7 A Minimization Example—The Profit & Gambit Co. Advertising-Mix Problem 47
FIGURE 2.19
The Engine tab of the
Model pane after selecting
the Standard LP/­Quadratic
Engine and setting the
Assume Nonnegative
option to True.
FIGURE 2.20
A screenshot showing
the final solution of the
Wyndor problem and the
Output tab of the model
pane showing a summary
of the solution process.
Once the model is all set up in Analytic Solver, the model would be solved by clicking on the
Optimize button on the Analytic Solver ribbon or the green arrow button on the Model pane.
Just like Excel’s Solver, this will then display the results of solving the model on the spreadsheet, as shown in Figure 2.20. As seen in this figure, the Output tab of the Model pane also
will show a summary of the solution process, including the message (similar to Figure 2.11)
that “Solver found a solution. All constraints and optimality conditions are satisfied.”
Review
Questions
1.
2.
3.
4.
Which button on the Analytic Solver ribbon should be pressed to specify the objective cell?
Which button on the Analytic Solver ribbon should be pressed to specify the changing cells?
Which button on the Analytic Solver ribbon should be pressed to enter the constraints?
Which button on the Analytic Solver ribbon should be pressed to solve the model?
2.7 A MINIMIZATION EXAMPLE—THE PROFIT & GAMBIT CO.
ADVERTISING-MIX PROBLEM
The analysis of the Wyndor Glass Co. case study in Sections 2.2, 2.5, and 2.6 illustrated
how to formulate and solve one type of linear programming model on a spreadsheet. The
same general approach can be applied to many other problems as well. The great flexibility
of linear programming and spreadsheets provides a variety of options for how to adapt the
48 Chapter Two Linear Programming: Basic Concepts
formulation of the spreadsheet model to fit each new problem. Our next example illustrates
some options not used for the Wyndor problem.
Planning an Advertising Campaign
The Profit & Gambit Co. produces cleaning products for home use. This is a highly competitive market, and the company continually struggles to increase its small market share.
Management has decided to undertake a major new advertising campaign that will focus on
the following three key products:
• A spray prewash stain remover.
• A liquid laundry detergent.
• A powder laundry detergent.
This campaign will use both television and the print media. A commercial has been developed
to run on national television that will feature the liquid detergent. The advertisement for the print
media will promote all three products and will include cents-off coupons that consumers can use
to purchase the products at reduced prices. The general goal is to increase the sales of each of
these products (but especially the liquid detergent) over the next year by a significant percentage
over the past year. Specifically, management has set the following goals for the campaign:
• Sales of the stain remover should increase by at least 3 percent.
• Sales of the liquid detergent should increase by at least 18 percent.
• Sales of the powder detergent should increase by at least 4 percent.
Table 2.2 shows the estimated increase in sales for each unit of advertising in the respective
outlets.4 (A unit is a standard block of advertising that Profit & Gambit commonly purchases,
but other amounts also are allowed.) The reason for −1 percent for the powder detergent in the
Television column is that the TV commercial featuring the new liquid detergent will take away
some sales from the powder detergent. The bottom row of the table shows the cost per unit of
advertising for each of the two outlets.
Management’s objective is to determine how much to advertise in each medium to meet
the sales goals at a minimum total cost.
Formulating a Spreadsheet Model for This Problem
The procedure summarized at the end of Section 2.2 can be used to formulate the spreadsheet
model for this problem. Each step of the procedure is repeated below, followed by a description of how it is performed here.
1. Gather the data for the problem. This has been done as presented in Table 2.2.
2. Enter the data into data cells on a spreadsheet. The top half of Figure 2.21 shows this
spreadsheet. The data cells are in columns C and D (rows 4 and 8 to 10), as well as in cells
G8:G10. Note how this particular formatting of the spreadsheet has facilitated a direct
transfer of the data from Table 2.2.
TABLE 2.2
Increase in Sales per
Unit of Advertising
Data for the Profit &
Gambit Co. AdvertisingMix Problem
Product
Stain remover
Liquid detergent
Powder detergent
Unit cost
Television
0%
3
–1
$1 million
Print Media
1%
2
4
Minimum
Required Increase
3%
18
4
$2 million
4
A simplifying assumption is being made that each additional unit of advertising in a particular outlet will
yield the same increase in sales regardless of how much advertising already is being done. This becomes a
poor assumption when the levels of advertising under consideration can reach a saturation level (as in Case
8.1), but is a reasonable approximation for the small levels of advertising being considered in this problem.
2.7 A Minimization Example—The Profit & Gambit Co. Advertising-Mix Problem 49
FIGURE 2.21
The spreadsheet model
for the Profit & Gambit
problem, including the
formulas for the objective cell TotalCost (G14)
and the other output cells
in column E, as well as
the specifications needed
to set up Solver. The
changing cells, AdvertisingUnits (C14:D14),
show the optimal solution
obtained by Solver.
A
1
B
C
D
E
F
G
Profit & Gambit Co. Advertising-Mix Problem
2
3
4
Unit Cost ($millions)
Television
Print Media
1
2
5
6
Increased
Minimum
Sales
Increase
Increase in Sales per Unit of Advertising
7
8
Stain Remover
0%
1%
3%
≥
3%
9
Liquid Detergent
3%
2%
18%
≥
18%
10
Powder Detergent
–1%
4%
8%
≥
4%
11
12
Total Cost
13
14
Advertising Units
Solver Parameters
Set Objective Cell: TotalCost
To: Min
By Changing (Variable) Cells:
AdvertisingUnits
Subject to the Constraints:
IncreasedSales >= MinimumIncrease
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
Television
Print Media
($millions)
4
3
10
E
6
Increased
7
8
9
10
Sales
=SUMPRODUCT(C8:D8, AdvertisingUnits)
=SUMPRODUCT(C9:D9, AdvertisingUnits)
=SUMPRODUCT(C10:D10, AdvertisingUnits)
G
12
Total Cost
13
($millions)
14
=SUMPRODUCT(UnitCost, AdvertisingUnits)
Range Name
AdvertisingUnits
IncreasedSales
IncreasedSalesPerUnitAdvertising
MinimumIncrease
TotalCost
UnitCost
Cells
C14: D14
E8: E10
C8: D10
G8: G10
G14
C4: D4
3. Identify the decisions to be made on the levels of activities and designate changing cells
for making these decisions. In this case, the activities of concern are advertising on television and advertising in the print media, so the levels of these activities refer to the amount
of advertising in these media. Therefore, the decisions to be made are
Decision 1: TV = Number of units of advertising on television
​​
​  ​
​
​​
Decision 2: PM = Number of units of advertising in the print media
The two gray cells with light borders in Figure 2.21—C14 and D14—have been designated
as the changing cells to hold these numbers:
​​TV → cell C14​
PM → cell D14​​
with AdvertisingUnits as the range name for these cells. (See the bottom of Figure 2.21 for
a list of all the range names.) These are natural locations for the changing cells, since each
one is in the column for the corresponding advertising medium. To get started, an arbitrary
trial solution (such as all zeroes) is entered into these cells. (Figure 2.21 shows the optimal
solution after having already applied Solver.)
50 Chapter Two Linear Programming: Basic Concepts
4. Identify the constraints on these decisions and introduce output cells as needed to specify these constraints. The three constraints imposed by management are the goals for the
increased sales for the respective products, as shown in the rightmost column of Table 2.2.
These constraints are
Unlike the Wyndor problem, we need to use ≥ signs
for these constraints.
Stain remover:
Total increase in sales ≥ 3%
detergent:
​​Liquid
​  ​  ​  Total increase in sales ≥ 18%
​​​ ​
Powder detergent: Total increase in sales ≥ 4%
The second and third columns of Table 2.2 indicate that the total increases in sales from
both forms of advertising are
Total for stain remover
= 1% of PM
Total
for liquid detergent
​​
​  ​  ​  = 3% of TV ​+ 2% of PM​ ​​
Total for powder detergent = − 1% of TV + 4% of PM
Unlike the Wyndor problem, the objective now is to
minimize the objective cell.
Consequently, since rows 8, 9, and 10 in the spreadsheet are being used to provide information about the three products, cells E8, E9, and E10 are introduced as output cells to
show the total increase in sales for the respective products. In addition, ≥ signs have been
entered in column F to remind us that the increased sales need to be at least as large as the
numbers in column G. (The use of ≥ signs here rather than ≤ signs is one key difference
from the spreadsheet model for the Wyndor problem in Figure 2.3.)
5. Choose the overall measure of performance to be entered into the objective cell. Management’s stated objective is to determine how much to advertise in each medium to meet the
sales goals at a minimum total cost. Therefore, the total cost of the advertising is entered
in the objective cell TotalCost (G14). G14 is a natural location for this cell since it is in
the same row as the changing cells. The bottom row of Table 2.2 indicates that the number
going into this cell is
​Cost = (​​ ​​$1 million​)​​​ TV + (​​ ​​$2 million​)​​​ PM → cell G14​
6. Use a SUMPRODUCT function to enter the appropriate value into each output cell (including the objective cell). Based on the above expressions for cost and total increases in sales,
the SUMPRODUCT functions needed here for the output cells are those shown under the
right side of the spreadsheet in Figure 2.21. Note that each of these functions involves the
relevant data cells and the changing cells, AdvertisingUnits (C14:D14).
This spreadsheet model is a linear programming model, since it possesses all the characteristics of such models enumerated in Section 2.2.
Applying Solver to This Model
The procedure for using Solver to obtain an optimal solution for this model is basically the
same as described in Section 2.5 (for Excel’s Solver) or Section 2.6 (for Analytic Solver).
The Solver parameters are shown below the left-hand side of the spreadsheet in Figure 2.21.
In addition to specifying the objective cell and changing cells, the constraints that IncreasedSales ≥ MinimumIncrease have been specified in this box by using the Add Constraint dialog
box. Since the objective is to minimize total cost, Min also has been selected. (This is in contrast to the choice of Max for the Wyndor problem.)
Two options are also specified at the bottom of the Solver Parameters box on the lower
left-hand side of Figure 2.21. The changing cells need nonnegativity constraints (specified in the main Solver dialog box in Excel’s Solver, or on the Engine tab of the Model
pane in Analytic Solver) because negative values of advertising levels are not possible
alternatives. Choosing the Simplex LP solving method in Excel’s Solver (or the Standard LP/Quadratic Engine in Analytic Solver) specifies that this is a linear programming
model.
After running Solver, the optimal solution shown in the changing cells of the spreadsheet
in Figure 2.21 is obtained.
An Application Vignette
Samsung Electronics Corp., Ltd. (SEC), is a leading merchant
of dynamic and static random access memory devices and
other advanced digital integrated circuits. Its site at Kiheung,
South Korea (probably the largest semiconductor fabrication
site in the world), fabricates more than 300,000 silicon wafers
per month and employs over 10,000 people.
Cycle time is the industry’s term for the elapsed time from
the release of a batch of blank silicon wafers into the fabrication
process until completion of the devices that are fabricated on
those wafers. Reducing cycle times is an ongoing goal since it
both decreases costs and enables offering shorter lead times
to potential customers, a real key to maintaining or increasing
market share in a very competitive industry.
Three factors present particularly major challenges when
striving to reduce cycle times. One is that the product mix
changes continually. Another is that the company often needs
to make substantial changes in the fab-out schedule inside the
target cycle time as it revises forecasts of customer demand.
The third is that the machines of a general type are not homogeneous so only a small number of machines are qualified to
perform each device step.
A management science team developed a huge linear programming model with tens of thousands of decision variables
and functional constraints to cope with these challenges. The
objective function involved minimizing back-orders and finished-goods inventory.
The ongoing implementation of this model enabled the company to reduce manufacturing cycle times to fabricate dynamic random access memory devices from more than 80 days to less than
30 days. This tremendous improvement and the resulting reduction in both manufacturing costs and sale prices enabled Samsung
to capture an additional $200 million in annual sales revenue.
Source: R. C. Leachman, J. Kang, and Y. Lin, “SLIM: Short Cycle Time
and Low Inventory in Manufacturing at Samsung Electronics,” Interfaces
32, no. 1 (January–February 2002), pp. 61–77. (A link to this article is
available at www.mhhe.com/Hillier6e.)
Optimal Solution
C14 = 4 (Undertake 4 units of advertising on television)
C14 = 3 (Undertake 3 units of advertising in the print media)
The objective cell indicates that the total cost of this advertising plan would be $10 million.
The Mathematical Model in the Spreadsheet
When performing step 5 of the procedure for formulating a spreadsheet model, the total cost
of advertising was determined to be
​Cost = TV + 2 PM (in millions of dollars)​
where the goal is to choose the values of TV (number of units of advertising on television)
and PM (number of units of advertising in the print media) so as to minimize this cost. Step 4
identified three functional constraints:
Stain remover:
1% of PM ≥ 3%
detergent:
​​Liquid
​  ​  ​    3% of TV + 2% of PM ≥ 18%
​​​ ​
Powder detergent:
−1% of TV + 4% of PM ≥ 4%
Choosing the Make Variables Non-Negative option with Solver recognized that TV and PM
cannot be negative. Therefore, after dropping the percentage signs from the functional constraints, the complete mathematical model in the spreadsheet can be stated in the following
succinct form.
​Minimize Cost = TV + 2 PM (in millions of dollars)​
subject to
Stain remover increased sales:
PM ≥ 3
​​Liquid detergent increased sales:
​  ​  ​  3 TV + 2 PM ≥ 18
​ ​​​
Powder detergent increased sales: − TV + 4 PM ≥ 4
and
​​TV ≥ 0​
PM ≥ 0​​
Implicit in this statement is “Choose the values of TV and PM so as to . . . .” The term “subject to” is shorthand for “Choose these values subject to the requirement that the values satisfy all the following constraints.”
51
52 Chapter Two Linear Programming: Basic Concepts
PM
FIGURE 2.22
Graph showing two objective function lines for
the Profit & Gambit Co.
advertising-mix problem,
where the bottom one
passes through the optimal solution.
10
Feasible
region
Cost = 15 = TV + 2 PM
Cost = 10 = TV + 2 PM
4
(4,3)
optimal
solution
0
10
5
Amount of TV advertising
15
TV
This model is the algebraic version of the linear programming model in the spreadsheet
(Figure 2.21). Note how the parameters (constants) of this algebraic model come directly
from the numbers in Table 2.2. In fact, the entire model could have been formulated directly
from this table.
The differences between this algebraic model and the one obtained for the Wyndor problem in Section 2.3 lead to some interesting changes in how the graphical method is applied to
solve the model. To further expand your geometric intuition about linear programming, we
briefly describe this application of the graphical method next.
Since this linear programming model has only two decision variables, it can be solved by
the graphical method described in Section 2.4. The method needs to be adapted in two ways
to fit this particular problem. First, because all the functional constraints now have a ≥ sign
with a positive right-hand side, after obtaining the constraint boundary lines in the usual way,
the arrows indicating which side of each line satisfies that constraint now all point away from
the origin. Second, the method is adapted to minimization by moving the objective function
lines in the direction that reduces Cost and then stopping at the last instant that an objective
function line still passes through a point in the feasible region, where such a point then is an
optimal solution. The supplement to this chapter includes a description of how the graphical
method is applied to the Profit & Gambit problem in this way.
Figure 2.22 shows the resulting final graph that identifies the optimal solution as
TV = 4
(use 4 units of TV advertising)
​​
​  ​
​
​​
PM = 3
(use 3 units of print media advertising)
Review
Questions
2.8
1.
2.
3.
4.
What kind of product is produced by the Profit & Gambit Co.?
Which advertising media are being considered for the three products under consideration?
What is management’s objective for the problem being addressed?
What was the rationale for the placement of the objective cell and the changing cells in the
spreadsheet model?
5. The algebraic form of the linear programming model for this problem differs from that for the
Wyndor Glass Co. problem in which two major ways?
LINEAR PROGRAMMING FROM A BROADER PERSPECTIVE
Linear programming is an invaluable aid to managerial decision making in all kinds of companies throughout the world. The emergence of powerful spreadsheet packages has helped
to further spread the use of this technique. The ease of formulating and solving small linear
programming models on a spreadsheet now enables some managers with a very modest background in management science to do this themselves on their own desktop.
2.8 Linear Programming from a Broader Perspective 53
Many linear programming studies are major projects involving decisions on the levels of
many hundreds or thousands of activities. For such studies, sophisticated software packages
that go beyond spreadsheets generally are used for both the formulation and solution processes. These studies normally are conducted by technically trained teams of management
scientists, sometimes called operations research analysts. Management instigates the study
and then clarifies its objectives and needs, but the team does the rest.
Consequently, there is little reason for a manager to know the details of how linear programming models are solved beyond the rudiments of using Solver. (Even most management
science teams will use commercial software packages for solving their models on a computer
rather than developing their own software.) Similarly, a manager does not need to know the
technical details of how to formulate complex models, how to validate such a model, how to
interact with the computer when formulating and solving a large model, how to efficiently
perform what-if analysis with such a model, and so forth. Therefore, these details are deemphasized in this book. A student who becomes interested in conducting technical analyses
as part of a management science team should plan to take additional, more technically oriented courses in management science.
So what does an enlightened manager need to know about linear programming? A manager needs to have a good intuitive feeling for what linear programming is. One objective of
this chapter is to begin to develop that intuition. That’s the purpose of studying the graphical
method for solving two-variable problems. It is rare to have a real linear programming problem with as few as two decision variables. Therefore, the graphical method has essentially no
practical value for solving real problems. However, it has great value for conveying the basic
notion that linear programming involves pushing up against constraint boundaries and moving objective function values in a favorable direction as far as possible. Chapter 14, available
at www.mhhe.com/Hillier6e, also demonstrates that this approach provides considerable
geometric insight into how to analyze larger models by other methods.
A manager must also have an appreciation for the relevance and power of linear programming to encourage its use where appropriate. For future managers using this book, this appreciation is being promoted by using application vignettes and their linked articles to describe
real applications of linear programming and the resulting impact, as well as by including (in
miniature form) various realistic examples and case studies that illustrate what can be done.
Certainly a manager must be able to recognize situations where linear programming is
applicable. We focus on developing this skill in Chapter 3, where you will learn how to recognize the identifying features for each of the major types of linear programming problems
(and their mixtures).
In addition, a manager should recognize situations where linear programming should
not be applied. Chapter 8 will help to develop this skill by examining certain underlying
assumptions of linear programming and the circumstances that violate these assumptions.
That chapter also describes other approaches that can be applied where linear programming
should not.
A manager needs to be able to distinguish between competent and shoddy studies using
linear programming (or any other management science technique). Therefore, another goal of
the upcoming chapters is to demystify the overall process involved in conducting a management science study, all the way from first studying a problem to final implementation of the
managerial decisions based on the study. This is one purpose of the case studies throughout
the book.
Finally, a manager must understand how to interpret the results of a linear programming
study. He or she especially needs to understand what kinds of information can be obtained
through what-if analysis, as well as the implications of such information for managerial decision making. Chapter 5 focuses on these issues.
Review
Questions
1. Does management generally get heavily involved with the technical details of a linear programming study?
2. What is the purpose of studying the graphical method for solving problems with two decision
variables when essentially all real linear programming problems have more than two?
3. List the things that an enlightened manager should know about linear programming.
54 Chapter Two Linear Programming: Basic Concepts
2.9 Summary
Linear programming is a powerful technique for aiding managerial decision making for certain kinds
of problems. The basic approach is to formulate a mathematical model called a linear programming
model to represent the problem and then to analyze this model. Any linear programming model includes
decision variables to represent the decisions to be made, constraints to represent the restrictions on the
feasible values of these decision variables, and an objective function that expresses the overall measure
of performance for the problem.
Spreadsheets provide a flexible and intuitive way of formulating and solving a linear programming
model. The data are entered into data cells. Changing cells display the values of the decision variables,
and an objective cell shows the value of the objective function. Output cells are used to help specify the
constraints. After formulating the model on the spreadsheet, Solver is used to quickly find an optimal
solution. Analytic Solver also provides a more powerful method for finding an optimal solution.
The graphical method can be used to solve a linear programming model having just two decision
variables. This method provides considerable insight into the nature of linear programming models and
optimal solutions.
Glossary
absolute reference A reference to a cell (or a
column or a row) with a fixed address, as indicated either by using a range name or by placing
a $ sign in front of the letter and number of the
cell reference. (Section 2.2 and Appendix A), 30
changing cells The cells in the spreadsheet that
show the values of the decision variables. (Section 2.2), 28
constraint A restriction on the feasible values of
the decision variables. (Sections 2.2 and 2.3), 32
constraint boundary equation The equation for
the constraint boundary line. (Section 2.4), 37
constraint boundary line For linear programming problems with two decision variables,
the line forming the boundary of the solutions
that are permitted by the constraint. (Section
2.4), 37
data cells The cells in the spreadsheet that show
the data of the problem. (Section 2.2), 28
decision variable An algebraic variable that
represents a decision regarding the level of a particular activity. The value of the decision variable
appears in a changing cell on the spreadsheet.
(Section 2.3), 34
feasible region The geometric region that consists of all the feasible solutions. (Section 2.4), 35
feasible solution A solution that simultaneously
satisfies all the constraints in the linear programming model. (Section 2.3), 34
functional constraint A constraint with a function of the decision variables on the left-hand
side. All constraints in a linear programming
model that are not nonnegativity constraints are
called functional constraints. (Section 2.3), 34
graphical method A method for solving linear
programming problems with two decision
variables on a two-dimensional graph.
(Section 2.4), 35
infeasible solution A solution that violates at
least one of the constraints in the linear programming model. (Section 2.3), 34
linear programming model The mathematical
model that represents a linear programming problem. (Sections 2.2 and 2.3), 28
nonnegativity constraint A constraint that
expresses the restriction that a particular decision
variable must be nonnegative (greater than or
equal to zero). (Section 2.3), 34
objective cell The cell in the spreadsheet that
shows the overall measure of performance of the
decisions. (Section 2.2), 30
objective function The part of a linear programming model that expresses what needs to be
either maximized or minimized, depending on the
objective for the problem. The value of the objective function appears in the objective cell on the
spreadsheet. (Section 2.3), 34
objective function line For a linear programming problem with two decision variables, a
line whose points all have the same value of the
objective function. (Section 2.4), 38
optimal solution The best feasible solution
according to the objective function.
(Section 2.3), 34
output cells The cells in the spreadsheet that
provide output that depends on the changing
cells. These cells frequently are used to help
specify constraints. (Section 2.2), 29
parameter The parameters of a linear programming model are the constants (coefficients or
right-hand sides) in the functional constraints
and the objective function. Each parameter represents a quantity (e.g., the amount available of a
resource) that is of importance for the analysis of
the problem. (Section 2.3), 34
product-mix problem A type of linear programming problem where the objective is to find the
most profitable mix of production levels for the
products under consideration. (Section 2.1), 27
range name A descriptive name given to a cell
or range of cells that immediately identifies what
is there. (Section 2.2), 28
Chapter 2 Problems 55
relative reference A reference to a cell whose
address is based upon its position relative to the
cell containing the formula. (Section 2.2 and
Appendix A), 30
solution Any single assignment of values to
the decision variables, regardless of whether the
assignment is a good one or even a feasible one.
(Section 2.3), 34
Solver The spreadsheet tool that is used to
specify the model in the spreadsheet and then
to obtain an optimal solution for that model.
(Section 2.5), 39
Learning Aids for This Chapter
All learning aids are available at www.mhhe.com/Hillier6e.
Excel Add-in:
Excel Files:
Analytic Solver
Wyndor Example
Supplement to This Chapter:
Profit & Gambit Example
More About the Graphical Method for Linear Programming
Solved Problems
The solutions are available at www.mhhe.com/Hillier6e.
2.S1. Back Savers Production Problem
Back Savers is a company that produces backpacks primarily
for students. They are considering offering some combination
of two different models—the Collegiate and the Mini. Both
are made out of the same rip-resistant nylon fabric. Back Savers has a long-term contract with a supplier of the nylon and
receives a 5,000-square-foot shipment of the material each week.
Each ­Collegiate requires 3 square feet while each Mini requires
2 square feet. The sales forecasts indicate that at most 1,000
Collegiates and 1,200 Minis can be sold per week. Each Collegiate requires 45 minutes of labor to produce and generates a
unit profit of $32. Each Mini requires 40 minutes of labor and
generates a unit profit of $24. Back Savers has 35 laborers that
each provides 40 hours of labor per week. Management wishes
to know what quantity of each type of backpack to produce per
week in order to maximize the total profit.
a. Formulate and solve a linear programming model for this
problem on a spreadsheet.
b. Formulate this same model algebraically.
c. Use the graphical method by hand to solve this model.
2.S2. Conducting a Marketing Survey
The marketing group for a cell phone manufacturer plans to conduct a telephone survey to determine consumer attitudes toward
a new cell phone that is currently under development. In order to
have a sufficient sample size to conduct the analysis, they need
to contact at least 100 young males (under age 40), 150 older
males (over age 40), 120 young females (under age 40), and
200 older females (over age 40). It costs $1 to make a daytime
phone call and $1.50 to make an evening phone call (because
of higher labor costs). This cost is incurred whether or not anyone answers the phone. The table below shows the likelihood of
a given customer type answering each phone call. Assume the
survey is conducted with whoever first answers the phone. Also,
because of limited evening staffing, at most one-third of phone
calls placed can be evening phone calls. How should the marketing group conduct the telephone survey so as to meet the sample
size requirements at the lowest possible cost?
a. Formulate and solve a linear programming model for this
problem on a spreadsheet.
b. Formulate this same model algebraically.
Who Answers?
Young male
Older male
Young female
Older female
No answer
Daytime Calls
Evening Calls
10%
15%
20%
35%
20%
20%
30%
20%
25%
5%
Problems
We have inserted the symbol AS to the left of each problem or
part where Analytic Solver is required. An asterisk on the problem number indicates that at least a partial answer is given in the
back of the book.
2.1. Read the referenced article that fully describes the management science study summarized in the application vignette
presented in Section 2.1. Briefly describe how linear programming was applied in this study. Then list the various financial
and nonfinancial benefits that resulted from this study.
2.2. Reconsider the Wyndor Glass Co. case study introduced
in Section 2.1. Suppose that the estimates of the unit profits for
the two new products now have been revised to $600 for the
doors and $300 for the windows.
a. Formulate and solve the revised linear programming model for this problem on a spreadsheet.
b. Formulate this same model algebraically.
c. Use the graphical method to solve this revised
model.
56 Chapter Two Linear Programming: Basic Concepts
2.3. Reconsider the Wyndor Glass Co. case study introduced
in Section 2.1. Suppose that Bill Tasto (Wyndor’s vice president
for manufacturing) now has found a way to provide a little additional production time in Plant 2 to the new products.
a. Use the graphical method to find the new optimal
solution and the resulting total profit if one additional hour per week is provided.
b. Repeat part a if two additional hours per week are
provided instead.
c. Repeat part a if three additional hours per week are
provided instead.
d. Use these results to determine how much each additional hour per week would be worth in terms of
increasing the total profit from the two new products.
2.4. Use Solver to do Problem 2.3.
2.5. The following table summarizes the key facts about two
products, A and B, and the resources, Q, R, and S, required to
produce them.
Resource Usage per
Unit Produced
Product
A
Product
B
Amount of
Resource
Available
Q
R
S
2
1
3
1
2
3
2
2
4
Profit/unit
$3,000
$2,000
Resource
All the assumptions of linear programming hold.
a. Formulate and solve a linear programming model
for this problem on a spreadsheet.
b. Formulate this same model algebraically.
2.6.* This is your lucky day. You have just won a $20,000 prize.
You are setting aside $8,000 for taxes and partying expenses,
but you have decided to invest the other $12,000. Upon hearing
this news, two different friends have offered you an opportunity
to become a partner in two different entrepreneurial ventures,
one planned by each friend. In both cases, this investment would
involve expending some of your time next summer as well as
putting up cash. Becoming a full partner in the first friend’s venture would require an investment of $10,000 and 400 hours, and
your estimated profit (ignoring the value of your time) would be
$9,000. The corresponding figures for the second friend’s venture are $8,000 and 500 hours, with an estimated profit to you of
$9,000. However, both friends are flexible and would allow you
to come in at any fraction of a full partnership you would like. If
you choose a fraction of a full partnership, all the above figures
given for a full partnership (money investment, time investment,
and your profit) would be multiplied by this same fraction.
Because you were looking for an interesting summer job anyway (maximum of 600 hours), you have decided to participate in
one or both friends’ ventures in whichever combination would
maximize your total estimated profit. You now need to solve the
problem of finding the best combination.
a. Describe the analogy between this problem and the
Wyndor Glass Co. problem discussed in Section
2.1. Then construct and fill in a table like Table 2.1
for this problem, identifying both the activities and
the resources.
b. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure
of performance for the decisions.
c. Convert these verbal descriptions of the constraints
and the measure of performance into quantitative
expressions in terms of the data and decisions.
d. Formulate a spreadsheet model for this problem.
Identify the data cells, the changing cells, and the
objective cell. Also show the Excel equation for
each output cell expressed as a SUMPRODUCT
function. Then use Solver to solve this model.
e. Indicate why this spreadsheet model is a linear programming model.
f. Formulate this same model algebraically.
g. Identify the decision variables, objective function,
nonnegativity constraints, functional constraints,
and parameters in both the algebraic version and
spreadsheet version of the model.
h. Use the graphical method by hand to solve this
model. What is your total estimated profit?
2.7. You are given the following linear programming model in
algebraic form, where x1 and x2 are the decision variables and Z
is the value of the overall measure of performance.
​​Maximize
​  Z = ​x​  1​​ + 2 ​x​  2​​
subject to
Constraint on resource 1: x1 + x2 ≤ 5 (amount available)
Constraint on resource 2: x1 + 3x2 ≤ 9 (amount available)
and
​​​x​  1​​ ≥ 0​
​x​  2​​ ≥ 0​​
a. Identify the objective function, the functional constraints, and the nonnegativity constraints in this
model.
b. Incorporate this model into a spreadsheet.
c. Is (x1, x2) = (3, 1) a feasible solution?
d. Is (x1, x2) = (1, 3) a feasible solution?
e. Use Solver to solve this model.
2.8. You are given the following linear programming model in
algebraic form, where x1 and x2 are the decision variables and Z
is the value of the overall measure of performance.
​​Maximize
​  Z = 3​x​  1​​ + 2​x​  2​​
subject to
Constraint on resource 1: 3x1 + x2 ≤ 9 (amount available)
Constraint on resource 2: x1 + 2x2 ≤ 8 (amount available)
and
​​​x​  1​​ ≥ 0​
​x​  2​​ ≥ 0​​
a. Identify the objective function, the functional constraints, and the nonnegativity constraints in this model.
b. Incorporate this model into a spreadsheet.
c. Is (x1, x2) = (2, 1) a feasible solution?
Chapter 2 Problems 57
d. Is (x1, x2) = (2, 3) a feasible solution?
e. Is (x1, x2) = (0, 5) a feasible solution?
f. Use Solver to solve this model.
2.9. The Whitt Window Company is a company with only three
employees that makes two different kinds of handcrafted windows: a wood-framed and an aluminum framed window. They
earn $60 profit for each wood-framed window and $30 profit for
each aluminum-framed window. Doug makes the wood frames
and can make 6 per day. Linda makes the aluminum frames and
can make 4 per day. Bob forms and cuts the glass and can make
48 square feet of glass per day. Each wood-framed window uses
6 square feet of glass and each aluminum-framed window uses 8
square feet of glass.
The company wishes to determine how many windows of
each type to produce per day to maximize total profit.
a. Describe the analogy between this problem and the
Wyndor Glass Co. problem discussed in Section
2.1. Then construct and fill in a table like Table 2.1
for this problem, identifying both the activities and
the resources.
b. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure
of performance for the decisions.
c. Convert these verbal descriptions of the constraints
and the measure of performance into quantitative
expressions in terms of the data and decisions.
d. Formulate a spreadsheet model for this problem.
Identify the data cells, the changing cells, and the
objective cell. Also show the Excel equation for
each output cell expressed as a SUMPRODUCT
function. Then use Solver to solve this model.
e. Indicate why this spreadsheet model is a linear programming model.
f. Formulate this same model algebraically.
g. Identify the decision variables, objective function,
nonnegativity constraints, functional constraints,
and parameters in both the algebraic version and
spreadsheet version of the model.
h. Use the graphical method to solve this model.
i. A new competitor in town has started making
wood-framed windows as well. This may force the
company to lower the price it charges and so lower
the profit made for each wood-framed window.
How would the optimal solution change (if at all) if
the profit per wood-framed window decreases from
$60 to $40? From $60 to $20?
j. Doug is considering lowering his working hours,
which would decrease the number of wood frames
he makes per day. How would the optimal solution
change if he only makes 5 wood frames per day?
2.10. The Apex Television Company has to decide on the
number of 65″ and 55″ sets to be produced at one of its factories.
Market research indicates that at most 280 of the 65″ sets and 70
of the 55″ sets can be sold per month. The maximum number of
work-hours available is 3,500 per month. A 65″ set requires 20
work-hours and a 55″ set requires 10 work-hours. Each 65″ set
sold produces a profit of $270 and each 55″ set produces a profit
of $180. A wholesaler has agreed to purchase all the television
sets produced if the numbers do not exceed the maxima indicated by the market research.
a. Formulate and solve a linear programming model
for this problem on a spreadsheet.
b. Formulate this same model algebraically.
c. Use the graphical method to solve this model.
2.11. The WorldLight Company produces two light fixtures
(Products 1 and 2) that require both metal frame parts and electrical components. Management wants to determine how many
units of each product to produce per week so as to maximize
profit. For each unit of Product 1, one unit of frame parts and
two units of electrical components are required. For each unit of
Product 2, three units of frame parts and two units of electrical
components are required. The company has a weekly supply of
3,000 units of frame parts and 4,500 units of electrical components. Each unit of Product 1 gives a profit of $13, and each unit
of Product 2, up to 900 units, gives a profit of $26. Any excess
over 900 units of Product 2 brings no profit, so such an excess
has been ruled out.
a. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure
of performance for the decisions.
b. Convert these verbal descriptions of the constraints
and the measure of performance into quantitative
expressions in terms of the data and decisions.
c. Formulate and solve a linear programming model
for this problem on a spreadsheet.
d. Formulate this same model algebraically.
2.12. The Primo Insurance Company is introducing two
new product lines: special risk insurance and mortgages. The
expected profit is $5 per unit on special risk insurance and $2
per unit on mortgages.
Management wishes to establish sales quotas for the new
product lines to maximize total expected profit. The work
requirements are shown below:
Work-Hours per Unit
Department
Special
Risk
Mortgage
Work-Hours
Available
Underwriting
Administration
Claims
3
0
2
2
1
0
2,400
800
1,200
a. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure
of performance for the decisions.
b. Convert these verbal descriptions of the constraints
and the measure of performance into quantitative
expressions in terms of the data and decisions.
c. Formulate and solve a linear programming model
for this problem on a spreadsheet.
d. Formulate this same model algebraically.
2.13.* You are given the following linear programming model
in algebraic form, with x1 and x2 as the decision variables and
constraints on the usage of four resources:
​​Maximize
​  Profit = 2​x​  1​​ + x​ ​  2​​
58 Chapter Two Linear Programming: Basic Concepts
subject to
and
​x​  2​​ ≤ 10
​​(​​resource 1​)​​​
2​x​  1​​ + 5​x​  2​​ ≤ 60
​​(​​resource 2​)​​​
​​
​
​​​
​x​  1​​ + ​x​  2​​ ≤ 18
​​(​​resource 3​)​​​
3​x​  1​​ + ​x​  2​​ ≤ 44
​​(​​resource 4​)​​​
​​​x​  1​​ ≥ 0​
​x​  2​​ ≥ 0​​
a. Use the graphical method to solve this model.
b. Incorporate this model into a spreadsheet and then
use Solver to solve this model.
AS 2.14. Use Analytic Solver to formulate and solve the model
shown in Problem 2.13 in a spreadsheet.
2.15. Because of your knowledge of management science, your
boss has asked you to analyze a product mix problem involving
two products and two resources. The model is shown below in
algebraic form, where x1 and x2 are the production rates for the
two products and P is the total profit.
​​Maximize
subject to
and
​  P = 3​x​  1​​ + 2​x​  2​​
​x​  1​​ + ​x​  2​​≤   8
​​(​​resource 1​)​​​
​​
​
​​​
2 ​x​  1​​ + ​x​  2​​ ≤ 10
​​(​​resource 2​)​​​
​​​x​  1​​ ≥ 0​
​x​  2​​ ≥ 0​​
a. Use the graphical method to solve this model.
b. Incorporate this model into a spreadsheet and then
use Solver to solve this model.
AS 2.16. Use Analytic Solver to formulate and solve the model
shown in Problem 2.15 in a spreadsheet.
2.17. Independently consider each of the following changes in
the Wyndor problem. In each case, apply the graphical method
by hand to this new version of the problem, describe your conclusion, and then explain how and why the nature of this conclusion is different from the original Wyndor problem.
a. The unit profit for the windows now is $200.
b. To justify introducing these two new products,
Wyndor management now requires that the total
number of doors and windows produced per week
must be at least 10.
c. The functional constraints for Plants 2 and 3 now
have been inadvertently deleted from the model.
2.18. Weenies and Buns is a food processing plant that manufactures buns and frankfurters for hot dogs. They grind their own
flour for the buns at a maximum rate of 200 pounds per week. Each
bun requires 0.1 pound of flour. They currently have a contract
Item
Beef tips
Gravy
Peas
Carrots
Dinner roll
with Pigland, Inc., which specifies that a delivery of 800 pounds of
pork product is delivered every Monday. Each frankfurter requires
1/4 pound of pork product. All the other ingredients in the buns
and frankfurters are in plentiful supply. Finally, the labor force at
Weenies and Buns consists of five employees working full time (40
hours per week each). Each bun requires two minutes of labor, and
each frankfurter requires three minutes of labor. Each bun yields a
profit of $0.20, and each frankfurter yields a profit of $0.40.
Weenies and Buns would like to know how many buns and
how many frankfurters they should produce each week so as to
achieve the highest possible profit.
a. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure
of performance for the decisions.
b. Convert these verbal descriptions of the constraints
and the measure of performance into quantitative
expressions in terms of the data and decisions.
c. Formulate and solve a linear programming model
for this problem on a spreadsheet.
d. Formulate this same model algebraically.
e. Use the graphical method to solve this model.
2.19. The Oak Works is a family-owned business that makes
handcrafted dining room tables and chairs. They obtain the oak
from a local tree farm, which ships them 2,500 pounds of oak each
month. Each table uses 50 pounds of oak while each chair uses
25 pounds of oak. The family builds all the furniture itself and
has 480 hours of labor available each month. Each table or chair
requires six hours of labor. Each table nets Oak Works $400 in
profit, while each chair nets $100 in profit. Since chairs are often
sold with the tables, they want to produce at least twice as many
chairs as tables.
The Oak Works would like to decide how many tables and
chairs to produce so as to maximize profit.
AS
a. Formulate and solve a linear programming model
for this problem on a spreadsheet by using the Excel
Solver.
b. Use Analytic Solver to formulate and solve this
model in a spreadsheet.
c. Formulate this same model algebraically.
2.20. Read the referenced article that fully describes the management science study summarized in the application vignette
presented in Section 2.6. Briefly describe how linear programming was applied in this study. Then list the various financial
and nonfinancial benefits that resulted from this study.
2.21. Nutri-Jenny is a weight-management center. It produces
a wide variety of frozen entrées for consumption by its clients. The entrées are strictly monitored for nutritional content
to ensure that the clients are eating a balanced diet. One new
Calories
(per oz.)
Calories
from Fat
(per oz.)
Vitamin A
(IU per oz.)
Vitamin C
(mg per oz.)
Protein
(g, per oz.)
Cost
(per oz.)
54
20
15
8
40
19
15
0
0
10
0
0
15
350
0
0
1
3
1
0
8
0
1
1
1
40¢
35¢
15¢
18¢
10¢
Chapter 2 Problems 59
entrée will be a beef sirloin tips dinner. It will consist of beef
tips and gravy, plus some combination of peas, carrots, and a
dinner roll. Nutri-Jenny would like to determine what quantity
of each item to include in the entrée to meet the nutritional
requirements, while costing as little as possible. The nutritional
information for each item and its cost are given in the table on
the bottom of the preceding page.
The nutritional requirements for the entrée are as follows:
(1) it must have between 280 and 320 calories, (2) calories
from fat should be no more than 30 percent of the total number
of calories, and (3) it must have at least 600 IUs of vitamin A,
10 milligrams of vitamin C, and 30 grams of protein. Furthermore, for practical reasons, it must include at least 2 ounces of
beef, and it must have at least half an ounce of gravy per ounce
of beef.
a. Formulate and solve a linear programming model
for this problem on a spreadsheet by using the Excel
Solver.
AS
b. Use Analytic Solver to formulate and solve this
model in a spreadsheet.
c. Formulate this same model algebraically.
2.22. Ralph Edmund loves steaks and potatoes. Therefore, he
has decided to go on a steady diet of only these two foods (plus
some liquids and vitamin supplements) for all his meals. Ralph
realizes that this isn’t the healthiest diet, so he wants to make
sure that he eats the right quantities of the two foods to satisfy
some key nutritional requirements. He has obtained the following nutritional and cost information.
Grams of Ingredient per Serving
Steak
Potatoes
Daily
Requirement
(grams)
Carbohydrates
Protein
Fat
5
20
15
15
5
2
≥ 50
≥ 40
≤60
Cost per serving
$4
$2
Ingredient
Ralph wishes to determine the number of daily servings (may
be fractional) of steak and potatoes that will meet these requirements at a minimum cost.
a. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure
of performance for the decisions.
b. Convert these verbal descriptions of the constraints
and the measure of performance into quantitative
expressions in terms of the data and decisions.
c. Formulate and solve a linear programming model
for this problem on a spreadsheet.
d. Formulate this same model algebraically.
e. Use the graphical method by hand to solve this model.
2.23. Dwight is an elementary school teacher who also
raises pigs for supplemental income. He is trying to decide
what to feed his pigs. He is considering using a combination
of pig feeds available from local suppliers. He would like to
feed the pigs at minimum cost while also making sure each
pig receives an adequate supply of calories and vitamins. The
cost, calorie content, and vitamin content of each feed is given
in the table below.
Contents
Feed Type A
Feed Type B
Calories (per pound)
Vitamins (per pound)
Cost (per pound)
800
140 units
$0.40
1,000
70 units
$0.80
Each pig requires at least 8,000 calories per day and at least
700 units of vitamins. A further constraint is that no more than
1/3 of the diet (by weight) can consist of Feed Type A, since it
contains an ingredient that is toxic if consumed in too large a
quantity.
a. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure
of performance for the decisions.
b. Convert these verbal descriptions of the constraints
and the measure of performance into quantitative
expressions in terms of the data and decisions.
c. Formulate and solve a linear programming model
for this problem on a spreadsheet.
d. Formulate this same model algebraically.
2.24. Reconsider the Profit & Gambit Co. problem described
in Section 2.7. Suppose that the estimated data given in Table
2.2 now have been changed as shown in the table that accompanies this problem.
a. Formulate and solve a linear programming model
on a spreadsheet for this revised version of the
problem.
b. Formulate this same model algebraically.
c. Use the graphical method to solve this model.
d. What were the key changes in the data that caused
your answer for the optimal solution to change from
the one for the original version of the problem?
Increase in Sales per
Unit of Advertising
Product
Television Print Media
Stain remover
Liquid detergent
Powder detergent
Unit cost
0%
3
−1
$1 million
1.5%
4
2
$2 million
Minimum
Required
Increase
3%
18
4
e. Write a paragraph to the management of the Profit
& Gambit Co. presenting your conclusions from the
above parts. Include the potential effect of further
refining the key data in the above table. Also point
out the leverage that your results might provide to
management in negotiating a decrease in the unit
cost for either of the advertising media.
2.25. You are given the following linear programming model
in algebraic form, with x1 and x2 as the decision variables:
​​Minimize
​
Cost = 40​x​  1​​ + 50​x​  2​​
60 Chapter Two Linear Programming: Basic Concepts
subject to
Constraint 1:
2​x​  1​​ + 3​x​  2​​ ≥ 30
2:​ ​
​x​  1​​ + ​ x​  2​​ ≥ 12​​​​
​​Constraint
Constraint 3:
2​x​  1​​ + ​ x​  2​​ ≥ 20
and
​​​x​  1​​ ≥ 0​
​x​  2​​ ≥ 0​​
a. Use the graphical method to solve this model.
b. How does the optimal solution change if the objective function is changed to Cost = 40x1 + 70x2?
c. How does the optimal solution change if the third
functional constraint is changed to 2x1 + x2 ≥ 15?
d. Now incorporate the original model into a spreadsheet and use Solver to solve this model.
e. Use Excel to do parts b and c.
2.26. The Learning Center runs a day camp for 6- to 10-yearolds during the summer. Its manager, Elizabeth Reed, is trying
to reduce the center’s operating costs to avoid having to raise
the tuition fee. Elizabeth is currently planning what to f eed
Food Item
Bread (1 slice)
Peanut butter (1 tbsp)
Jelly (1 tbsp)
Apple
Milk (1 cup)
Cranberry juice (1 cup)
the children for lunch. She would like to keep costs to a minimum, but also wants to make sure she is meeting the nutritional
requirements of the children. She has already decided to go with
peanut butter and jelly sandwiches, and some combination of
apples, milk, and/or cranberry juice. The nutritional content of
each food choice and its cost are given in the table that accompanies this problem.
The nutritional requirements are as follows. Each child
should receive between 300 and 500 calories, but no more than
30 percent of these calories should come from fat. Each child
should receive at least 60 milligrams (mg) of vitamin C and at
least 10 grams (g) of fiber.
To ensure tasty sandwiches, Elizabeth wants each child to
have a minimum of 2 slices of bread, 1 tablespoon (tbsp) of peanut butter, and 1 tbsp of jelly, along with at least 1 cup of liquid
(milk and/or cranberry juice).
Elizabeth would like to select the food choices that would
minimize cost while meeting all these requirements.
a. Formulate and solve a linear programming model
for this problem on a spreadsheet.
b. Formulate this same model algebraically.
Calories
from Fat
Total
Calories
Vitamin C
(mg)
Fiber (g)
Cost (¢)
15
80
0
0
60
0
80
100
70
90
120
110
0
0
4
6
2
80
4
0
3
10
0
1
6
5
8
35
20
40
Case 2-1
Auto Assembly
Automobile Alliance, a large automobile manufacturing company, organizes the vehicles it manufactures into three families: a
family of trucks, a family of small cars, and a family of midsized
and luxury cars. One plant outside Detroit, Michigan, assembles
two models from the family of midsized and luxury cars. The
first model, the Family Thrillseeker, is a four-door sedan with
vinyl seats, plastic interior, standard features, and excellent gas
mileage. It is marketed as a smart buy for middle-class families
with tight budgets, and each Family Thrillseeker sold generates
a modest profit of $3,600 for the company. The second model,
the Classy Cruiser, is a two-door luxury sedan with leather seats,
wooden interior, custom features, and navigational capabilities.
It is marketed as a privilege of affluence for upper-middle-class
families, and each Classy Cruiser sold generates a healthy profit
of $5,400 for the company.
Rachel Rosencrantz, the manager of the assembly plant, is
currently deciding the production schedule for the next month.
Specifically, she must decide how many Family Thrillseekers
and how many Classy Cruisers to assemble in the plant to maximize profit for the company. She knows that the plant possesses a
capacity of 48,000 labor-hours during the month. She also knows
that it takes six labor-hours to assemble one Family Thrillseeker
and 10.5 labor-hours to assemble one Classy Cruiser.
Because the plant is simply an assembly plant, the parts
required to assemble the two models are not produced at the
plant. Instead, they are shipped from other plants around the
Michigan area to the assembly plant. For example, tires, steering
wheels, windows, seats, and doors all arrive from various supplier plants. For the next month, Rachel knows that she will only
be able to obtain 20,000 doors from the door supplier. A recent
labor strike forced the shutdown of that particular supplier plant
for several days, and that plant will not be able to meet its production schedule for the next month. Both the Family Thrillseeker
and the Classy Cruiser use the same door parts, with four needed
for the Family Thrillseeker and two for the Classy Cruiser.
In addition, a recent company forecast of the monthly
demands for different automobile models suggests that the
demand for the Classy Cruiser is limited to 3,500 cars. There
is no limit on the demand for the Family Thrillseeker within the
capacity limits of the assembly plant.
Case 2-2 Cutting Cafeteria Costs 61
a. Formulate and solve a linear programming model to determine the number of Family Thrillseekers and the number of
Classy Cruisers that should be assembled.
Before she makes her final production decisions, Rachel plans
to explore the following questions independently, except where
otherwise indicated.
b. The marketing department knows that it can pursue a targeted
$500,000 advertising campaign that will raise the demand
for the Classy Cruiser next month by 20 percent. Should the
campaign be undertaken?
c. Rachel knows that she can increase next month’s plant capacity by using overtime labor. She can increase the plant’s
labor-hour capacity by 25 percent. With the new assembly
plant capacity, how many Family Thrillseekers and how
many Classy Cruisers should be assembled?
d. Rachel knows that overtime labor does not come without an
extra cost. What is the maximum amount she should be willing to pay for all overtime labor beyond the cost of this labor
at regular-time rates? Express your answer as a lump sum.
e. Rachel explores the option of using both the targeted advertising campaign and the overtime labor-hours. The advertising campaign raises the demand for the Classy Cruiser by
20 percent, and the overtime labor increases the plant’s laborhour capacity by 25 percent. How many Family Thrillseekers and how many Classy Cruisers should be assembled
using the advertising campaign and overtime labor-hours
if the profit from each Classy Cruiser sold continues to be
50 ­percent more than for each Family Thrillseeker sold?
f. Knowing that the advertising campaign costs $500,000 and
the maximum usage of overtime labor-hours costs $1,600,000
beyond regular time rates, is the solution found in part e a
wise decision compared to the solution found in part a?
g. Automobile Alliance has determined that dealerships are
actually heavily discounting the price of the Family Thrillseekers to move them off the lot. Because of a profit-­sharing
agreement with its dealers, the company is not making a
profit of $3,600 on the Family Thrillseeker but instead is
making a profit of $2,800. Determine the number of Family
Thrillseekers and the number of Classy Cruisers that should
be assembled given this new discounted profit.
h. The company has discovered quality problems with the
­Family Thrillseeker by randomly testing Thrillseekers at the
end of the assembly line. Inspectors have discovered that in
over 60 percent of the cases, two of the four doors on a Thrillseeker do not seal properly. Because the percentage of defective Thrillseekers determined by the random testing is so high,
the floor foreman has decided to perform quality control tests
on every Thrillseeker at the end of the line. Because of the
added tests, the time it takes to assemble one Family Thrillseeker has increased from 6 hours to 7.5 hours. Determine
the number of units of each model that should be assembled
given the new assembly time for the Family Thrillseeker.
i. The board of directors of Automobile Alliance wishes to capture a larger share of the luxury sedan market and therefore
would like to meet the full demand for Classy Cruisers. They
ask Rachel to determine by how much the profit of her assembly plant would decrease as compared to the profit found in
part a. They then ask her to meet the full demand for Classy
Cruisers if the decrease in profit is not more than $2,000,000.
j. Rachel now makes her final decision by combining all the
new considerations described in parts f, g, and h. What are
her final decisions on whether to undertake the advertising
campaign, whether to use overtime labor, the number of
Family Thrillseekers to assemble, and the number of Classy
Cruisers to assemble?
Case 2-2
Cutting Cafeteria Costs
A cafeteria at All-State University has one special dish it serves
like clockwork every Thursday at noon. This supposedly tasty dish
is a casserole that contains sautéed onions, boiled sliced potatoes,
green beans, and cream of mushroom soup. Unfortunately, students fail to see the special quality of this dish, and they loathingly
refer to it as the Killer Casserole. The students reluctantly eat the
casserole, however, because the cafeteria provides only a limited
selection of dishes for Thursday’s lunch (namely, the casserole).
Maria Gonzalez, the cafeteria manager, is looking to cut
costs for the coming year, and she believes that one sure way
to cut costs is to buy less expensive and perhaps lower quality
ingredients. Because the casserole is a weekly staple of the cafeteria menu, she concludes that if she can cut costs on the ingredients purchased for the casserole, she can significantly reduce
overall cafeteria operating costs. She therefore decides to invest
time in determining how to minimize the costs of the casserole
while maintaining nutritional and taste requirements.
Maria focuses on reducing the costs of the two main ingredients in the casserole, the potatoes and green beans. These two
ingredients are responsible for the greatest costs, nutritional content, and taste of the dish.
Maria buys the potatoes and green beans from a wholesaler
each week. Potatoes cost $0.40 per pound (lb), and green beans
cost $1.00 per lb.
All-State University has established nutritional requirements
that each main dish of the cafeteria must meet. Specifically, the
dish must contain 180 grams (g) of protein, 80 milligrams (mg)
of iron, and 1,050 mg of vitamin C. (There are 454 g in one lb
and 1,000 mg in one g.) For simplicity when planning, Maria
assumes that only the potatoes and green beans contribute to the
nutritional content of the casserole.
Because Maria works at a cutting-edge technological university,
she has been searching for information on the Internet to find the
nutritional content of potatoes and green beans. Her research yields
the following nutritional information about the two ingredients.
Protein
Iron
Vitamin C
Potatoes
Green Beans
1.5 g per 100 g
0.3 mg per 100 g
12 mg per 100 g
5.67 g per 10 ounces
3.402 mg per 10 ounces
28.35 mg per 10 ounces
(There are 28.35 g in one ounce.)
62 Chapter Two Linear Programming: Basic Concepts
Edson Branner, the cafeteria cook who is surprisingly concerned about taste, informs Maria that an edible casserole must
contain at least a six-to-five ratio in the weight of potatoes to
green beans.
Given the number of students who eat in the cafeteria, Maria
knows that she must purchase enough potatoes and green beans
to prepare a minimum of 10 kilograms (kg) of casserole each
week. (There are 1,000 g in one kg.) Again, for simplicity in
planning, she assumes that only the potatoes and green beans
determine the amount of casserole that can be prepared. Maria
does not establish an upper limit on the amount of casserole to
prepare since she knows all leftovers can be served for many days
thereafter or can be used creatively in preparing other dishes.
a. Determine the amount of potatoes and green beans Maria
should purchase each week for the casserole to minimize the
ingredient costs while meeting nutritional, taste, and demand
requirements.
Before she makes her final decision, Maria plans to explore
the following questions independently, except where otherwise
indicated.
b. Maria is not very concerned about the taste of the casserole;
she is only concerned about meeting nutritional requirements
and cutting costs. She therefore forces Edson to change the
recipe to allow only for at least a one-to-two ratio in the
weight of potatoes to green beans. Given the new recipe,
determine the amount of potatoes and green beans Maria
should purchase each week.
c. Maria decides to lower the iron requirement to 65 mg since
she determines that the other ingredients, such as the onions
and cream of mushroom soup, also provide iron. Determine
the amount of potatoes and green beans Maria should purchase each week given this new iron requirement.
d. Maria learns that the wholesaler has a surplus of green
beans and is therefore selling the green beans for a lower
price of $0.50 per lb. Using the same iron requirement
from part c and the new price of green beans, determine the
amount of potatoes and green beans Maria should purchase
each week.
e. Maria decides that she wants to purchase lima beans instead
of green beans since lima beans are less expensive and
provide a greater amount of protein and iron than green
beans. Maria again wields her absolute power and forces
Edson to change the recipe to include lima beans instead of
green beans. Maria knows she can purchase lima beans for
$0.60 per lb from the wholesaler. She also knows that lima
beans contain 22.68 g of protein and 6.804 mg of iron per
10 ounces of lima beans and no vitamin C. Using the new
cost and nutritional content of lima beans, determine the
amount of potatoes and lima beans Maria should purchase
each week to minimize the ingredient costs while meeting
nutritional, taste, and demand requirements. The nutritional
requirements include the reduced iron requirement from
part c.
f. Will Edson be happy with the solution in part e? Why or why
not?
g. An All-State student task force meets during Body Awareness Week and determines that All-State University’s nutritional requirements for iron are too lax and that those for
vitamin C are too stringent. The task force urges the university to adopt a policy that requires each serving of an entrée
to contain at least 120 mg of iron and at least 500 mg of vitamin C. Using potatoes and lima beans as the ingredients for
the dish and using the new nutritional requirements, determine the amount of potatoes and lima beans Maria should
purchase each week.
Case 2-3
Staffing a Call Center
California Children’s Hospital has been receiving numerous
customer complaints because of its confusing, decentralized
appointment and registration process. When customers want to
make appointments or register child patients, they must contact
the clinic or department they plan to visit. Several problems
exist with this current strategy. Parents do not always know
the most appropriate clinic or department they must visit to
address their children’s ailments. They therefore spend a significant amount of time on the phone being transferred from
clinic to clinic until they reach the most appropriate clinic for
their needs. The hospital also does not publish the phone numbers of all clinics and departments, and parents must therefore
invest a large amount of time in detective work to track down
the correct phone number. Finally, the various clinics and
departments do not communicate with each other. For example,
when a doctor schedules a referral with a colleague located in
another department or clinic, that department or clinic almost
never receives word of the referral. The parent must contact
the correct department or clinic and provide the needed referral
information.
In efforts to reengineer and improve its appointment and registration process, the children’s hospital has decided to centralize
the process by establishing one call center devoted exclusively
to appointments and registration. The hospital is currently in the
middle of the planning stages for the call center. Lenny Davis,
the hospital manager, plans to operate the call center from 7 AM
to 9 PM during the weekdays.
Several months ago, the hospital hired an ambitious management consulting firm, Creative Chaos Consultants, to forecast
the number of calls the call center would receive each hour of the
day. Since all appointment and registration-related calls would
be received by the call center, the consultants decided that they
could forecast the calls at the call center by totaling the number
of appointment and registration-related calls received by all clinics and departments. The team members visited all the clinics
and departments, where they diligently recorded every call relating to appointments and registration. They then totaled these
calls and altered the totals to account for calls missed during data
collection. They also altered totals to account for repeat calls that
occurred when the same parent called the hospital many times
Case 2-3 Additional Case 63
because of the confusion surrounding the decentralized process.
Creative Chaos Consultants determined the average number of
calls the call center should expect during each hour of a weekday. The following table provides the forecasts.
Work Shift
Average Number of Calls
7 AM to 9 AM
9 AM to 11 AM
11 AM to 1 PM
1 PM to 3 PM
3 PM to 5 PM
5 PM to 7 PM
7 PM to 9 PM
40 calls per hour
85 calls per hour
70 calls per hour
95 calls per hour
80 calls per hour
35 calls per hour
10 calls per hour
After the consultants submitted these forecasts, Lenny
became interested in the percentage of calls from Spanish speakers since the hospital services many Spanish patients. Lenny
knows that he has to hire some operators who speak Spanish to
handle these calls. The consultants performed further data collection and determined that, on average, 20 percent of the calls
were from Spanish speakers.
Given these call forecasts, Lenny must now decide how to
staff the call center during each two-hour shift of a week-day.
During the forecasting project, Creative Chaos Consultants
closely observed the operators working at the individual clinics
and departments and determined the number of calls operators
process per hour. The consultants informed Lenny that an operator is able to process an average of six calls per hour. Lenny also
knows that he has both full-time and part-time workers available
to staff the call center. A full-time employee works eight hours
per day, but because of paperwork that must also be completed,
the employee spends only four hours per day on the phone. To
balance the schedule, the employee alternates the two-hour shifts
between answering phones and completing paperwork. Fulltime employees can start their day either by answering phones
or by completing paperwork on the first shift. The full-time
employees speak either Spanish or English, but none of them are
bilingual. Both Spanish-speaking and English-speaking employees are paid $15 per hour for work before 5 PM and $18 per hour
for work after 5 PM. The full-time employees can begin work
at the beginning of the 7 AM to 9 AM shift, 9 AM to 11 AM shift,
11 AM to 1 PM shift, or 1 PM to 3 PM shift. The part-time employees work for four hours, only answer calls, and only speak
­English. They can start work at the beginning of the 3 PM to 5 PM
shift or the 5 PM to 7 pm shift, and, like the full-time employees,
they are paid $15 per hour for work before 5 PM and $18 per hour
for work after 5 PM.
For the following analysis, consider only the labor cost for
the time employees spend answering phones. The cost for paperwork time is charged to other cost centers.
a. How many Spanish-speaking operators and how many
­English-speaking operators does the hospital need to staff the
call center during each two-hour shift of the day in order to
answer all calls? Please provide an integer number since half
a human operator makes no sense.
b. Lenny needs to determine how many full-time employees
who speak Spanish, full-time employees who speak English,
and part-time employees he should hire to begin on each
shift. Creative Chaos Consultants advises him that linear programming can be used to do this in such a way as to minimize
operating costs while answering all calls. Formulate a linear
programming model of this problem.
c. Obtain an optimal solution for the linear programming model
formulated in part b to guide Lenny’s decision.
d. Because many full-time workers do not want to work late
into the evening, Lenny can find only one qualified Englishspeaking operator willing to begin work at 1 PM. Given this
new constraint, how many full-time English-speaking operators, full-time Spanish-speaking operators, and part-time
operators should Lenny hire for each shift to minimize operating costs while answering all calls?
e. Lenny now has decided to investigate the option of hiring
bilingual operators instead of monolingual operators. If all
the operators are bilingual, how many operators should be
working during each two-hour shift to answer all phone
calls? As in part a, please provide an integer answer.
f. If all employees are bilingual, how many full-time and parttime employees should Lenny hire to begin on each shift
to minimize operating costs while answering all calls? As
in part b, formulate a linear programming model to guide
­Lenny’s decision.
g. What is the maximum percentage increase in the hourly wage
rate that Lenny can pay bilingual employees over monolingual employees without increasing the total operating costs?
h. What other features of the call center should Lenny explore
to improve service or minimize operating costs?
Source: This case is based on an actual project completed by
a team of master’s students in what is now the Department of
Management Science and Engineering at Stanford University.
Additional Case
Additional cases for this chapter are also available at the University of Western Ontario Ivey School of Business website, cases.
ivey.uwo.ca/cases, in the segment of the Case-Mate area designated for this book.
Chapter Three
Linear Programming:
Formulation and
Applications
Learning Objectives
After completing this chapter, you should be able to
1. Recognize various kinds of managerial problems to which linear programming can be applied.
2. Describe the five major categories of linear programming problems, including their identifying features.
3. Formulate a linear programming model from a description of a problem in any of these
categories.
4. Describe the difference between resource constraints and benefit constraints, including the
difference in how they arise.
5. Describe fixed-requirement constraints and where they arise.
6. Identify the kinds of Excel functions that linear programming spreadsheet models use for
the output cells, including the objective cell.
7. Identify the four components of any linear programming model and the kind of spreadsheet
cells used for each component.
8. Understand the flexibility that managers have in prescribing key considerations that can be
incorporated into a linear programming model.
Chapter 2 introduced the basic concepts of linear programming, but only considered two tiny
examples (the Wyndor case study introduced in Section 2.1 and the Profit & Gambit problem
considered in Section 2.7) that only begin to illustrate the great power and versatility of this
technique. This chapter will expand considerably further on the great application potential of
linear programming by describing and illustrating several different types of linear programming problems. The many examples also will be supplemented by five solved problems presented at the end of the chapter.
Linear programming problems come in many guises. And their models take various forms.
This diversity can be confusing to both students and managers, making it difficult to recognize
when linear programming can be applied to address a managerial problem. Since managers
instigate management science studies, the ability to recognize the applicability of linear programming is an important managerial skill. This chapter focuses largely on developing this skill.
The usual textbook approach to trying to teach this skill is to present a series of diverse
examples of linear programming applications. The weakness of this approach is that it emphasizes differences rather than the common threads between these applications. Our approach
will be to emphasize these common threads—the identifying features—that tie together linear programming problems even when they arise in very different contexts. We will describe
some broad categories of linear programming problems and the identifying features that characterize them. Then we will use diverse examples, but with the purpose of illustrating and
emphasizing the common threads among them.
64
3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem 65
We will focus on the following five key categories of linear programming problems:
•
•
•
•
•
resource-allocation problems
cost-benefit-trade-off problems
mixed problems
transportation problems
assignment problems
In each case, an important identifying feature is the nature of the restrictions on what decisions can be made, and thus the nature of the resulting functional constraints in the linear
programming model. For each category, you will see how the basic data for a problem lead
directly to a linear programming model with a certain distinctive form. Thus, model formulation becomes a by-product of proper problem formulation.
The chapter begins with a case study (Section 3.1) that will illustrate the systematic procedure for formulating and solving the model for any linear programming problem. Section 3.2
will point out why the initial version of the case study turns out to be one example of a
resource-allocation problem. However, we then return to the case study in Section 3.4, where
additional managerial considerations turn the problem into a mixed problem.
Sections 3.2 to 3.6 focus on the five categories of linear programming problems in turn. (Section 3.4 also briefly introduces still another category, fixed-requirements problems, where this
category provides the framework for the more specific categories considered in Sections 3.5
and 3.6.) Section 3.7 then takes a broader look at the formulation of linear programming
models from a managerial perspective. That section (along with Section 3.4) highlights the
importance of having the model accurately reflect the managerial view of the problem. These
(and other) sections also describe the flexibility available to managers for having the model
structured to best fit their view of the important considerations.
3.1 A CASE STUDY: THE SUPER GRAIN CORP. ADVERTISING-MIX PROBLEM
Claire Syverson, vice president for marketing of the Super Grain Corporation, is facing a
daunting challenge: how to break into an already overly crowded breakfast cereal market in
a big way. Fortunately, the company’s new breakfast cereal—Crunchy Start—has a lot going
for it: Great taste. Nutritious. Crunchy from start to finish. She can recite the litany in her
sleep now. It has the makings of a winning promotional campaign.
However, Claire knows that she has to avoid the mistakes she made in her last campaign for
a breakfast cereal. That had been her first big assignment since she won this promotion, and
what a disaster! She thought she had developed a really good campaign. But somehow it had
failed to connect with the most crucial segments of the market—young children and parents
of young children. She also has concluded that it was a mistake not to include cents-off coupons in the magazine and newspaper advertising. Oh well. Live and learn.
But she had better get it right this time, especially after the big stumble last time. The company’s president, David Sloan, already has impressed on her how important the success of
Crunchy Start is to the future of the company. She remembers exactly how David concluded
the conversation. “The company’s shareholders are not happy. We need to get those earnings
headed in the right direction again.” Claire had heard this tune before, but she saw in David’s
eyes how deadly serious he was this time.
Claire often uses spreadsheets to help organize her planning. Her management science
course in business school impressed upon her how valuable spreadsheet modeling can be. She
regrets that she did not rely more heavily on spreadsheet modeling for the last campaign. That
was a mistake that she is determined not to repeat.
Now it is time for Claire to carefully review and formulate the problem in preparation for
formulating a spreadsheet model.
The Problem
Claire already has employed a leading advertising firm, Giacomi & Jackowitz, to help design
a nationwide promotional campaign that will achieve the largest possible exposure for Crunchy
66 Chapter Three Linear Programming: Formulation and Applications
TABLE 3.1
Costs
Cost and Exposure Data
for the Super Grain
Corp. Advertising-Mix
Problem
Cost Category
Each TV
Commercial
Each Magazine Ad
Each Sunday Ad
Ad budget
Planning budget
$300,000
90,000
$150,000
30,000
$100,000
40,000
Expected number of exposures
1,300,000
600,000
500,000
Start. Super Grain will pay this firm a fee based on services performed (not to exceed $1 million)
and has allocated an additional $4 million for advertising expenses.
Giacomi & Jackowitz has identified the three most effective advertising media for this product:
Medium 1: Television commercials on Saturday morning programs for children.
Medium 2: Advertisements in food and family-oriented magazines.
Medium 3: Advertisements in Sunday supplements of major newspapers.
The problem now is to determine which levels should be chosen for these advertising activities
to obtain the most effective advertising mix.
To determine the best mix of activity levels for this particular advertising problem, it is
necessary (as always) to identify the overall measure of performance for the problem and
then the contribution of each activity toward this measure. An ultimate goal for Super Grain
is to maximize its profits, but it is difficult to make a direct connection between advertising
exposure and profits. Therefore, as a rough surrogate for profit, Claire decides to use expected
number of exposures as the overall measure of performance, where each viewing of an advertisement by some individual counts as one exposure.
Giacomi & Jackowitz has made preliminary plans for advertisements in the three media.
The firm also has estimated the expected number of exposures for each advertisement in each
medium, as given in the bottom row of Table 3.1.
The number of advertisements that can be run in the different media are restricted by both
the advertising budget (a limit of $4 million) and the planning budget (a limit of $1 million
for the fee to Giacomi & Jackowitz). Another restriction is that there are only five commercial
spots available for running different commercials (one commercial per spot) on children’s
television programs Saturday morning (medium 1) during the time of the promotional campaign. (The other two media have an ample number of spots available.)
Consequently, the three resources for this problem are:
Resource 1: Advertising budget ($4 million).
Resource 2: Planning budget ($1 million).
Resource 3: TV commercial spots available (5).
Table 3.1 shows how much of the advertising budget and the planning budget would be
used by each advertisement in the respective media.
• The first row gives the cost per advertisement in each medium.
• The second row shows Giacomi & Jackowitz’s estimates of its total cost (including overhead and profit) for designing and developing each advertisement for the respective media.1
(This cost represents the billable fee from Super Grain.)
• The last row then gives the expected number of exposures per advertisement.
Analysis of the Problem
Claire decides to formulate and solve a linear programming model for this problem on a
spreadsheet. The formulation procedure summarized at the end of Section 2.2 guides this process. Like any linear programming model, this model will have four components:
1. The data
2. The decisions
1
When presenting its estimates in this form, the firm is making two simplifying assumptions. One is that its
cost for designing and developing each additional advertisement in a medium is roughly the same as for the
first advertisement in that medium. The second is that its cost when working with one medium is unaffected
by how much work it is doing (if any) with the other media.
3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem 67
3. The constraints
4. The measure of performance
The spreadsheet needs to be formatted to provide the following kinds of cells for these
components:
Data → data cells
Decisions → changing cells
Constraints → output cells
Measure of performance → objective cell
Four kinds of cells are
needed for these four components of a spreadsheet
model.
Figure 3.1 shows the spreadsheet model formulated by Claire. Let us see how she did this,
including why she structured the spreadsheet in this way, by considering each of the ­components
of the model individually.
FIGURE 3.1
The spreadsheet model for the Super Grain problem (Section 3.1), including the objective cell TotalExposures (H13) and the other
output cells BudgetSpent (F8:F9), as well as the specifications needed to set up Solver. The changing cells NumberOfAds (C13:E13)
show the optimal solution obtained by Solver.
A
1
B
D
E
TV Spots
Magazine Ads
SS Ads
1,300
600
500
C
F
G
H
Super Grain Corp. Advertising-Mix Problem
2
3
4
Exposures per Ad
5
(thousands)
6
Cost per Ad ($thousands)
7
Budget
Budget
Spent
Available
8
Ad Budget
300
150
100
4,000
≤
4,000
9
Planning Budget
90
30
40
1,000
≤
1,000
TV Spots
Magazine Ads
SS Ads
(thousands)
0
20
10
17,000
10
11
12
13
Number of Ads
14
15
Total Exposures
≤
Max TV Spots
Solver Parameters
Set Objective Cell: TotalExposures
To: Max
By Changing Variable Cells:
NumberOfAds
Subject to the Constraints:
BudgetSpent <= BudgetAvailable
TVSpots <= MaxTVSpots
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
5
F
6
Budget
7
8
9
Spent
=SUMPRODUCT(C8:E8,NumberOfAds)
=SUMPRODUCT(C9:E9,NumberOfAds)
H
11
Total Exposures
12
(thousands)
13
=SUMPRODUCT(ExposuresPerAd,NumberOfAds)
Range Name
Cells
BudgetAvailable
BudgetSpent
CostPerAd
ExposuresPerAd
MaxTVSpots
NumberOfAds
TotalExposures
TVSpots
H8: H9
F8: F9
C8: E9
C4: E4
C15
C13: E13
H13
C13
68 Chapter Three Linear Programming: Formulation and Applications
The Data
It is best to begin structuring the spreadsheet by deciding where to locate the data cells. Most
of these cells normally are placed in the upper left-hand portion of the spreadsheet. The structure of the rest of the model will then follow the structure of the data cells. When some or
all of the data are being obtained from a well-constructed table, the clarity of the spreadsheet usually is enhanced by retaining the format of the table when inserting the data into the
spreadsheet.
Claire followed these guidelines when she chose cells C4:E4 and C8:E9 to be the data cells
to hold the information in Table 3.1 when using units of thousands of dollars. These data cells
have been given the range names: ExposuresPerAd (C4:E4), CostPerAd (C8:E9), BudgetAvailable (H8:H9), and MaxTVSpots (C15).
The other relevant kind of data here is the information given earlier about the amounts
available of the three resources for the problem (the advertising budget, the planning budget,
and the commercial spots available). Claire chose the locations of the data cells holding this
information (H8:H9 and C15) to fit with the constraints that will be inserted a little later.
The Decisions
The problem has been defined as determining the most effective advertising mix among the
three media selected by Giacomi & Jackowitz. Therefore, there are three decisions:
Decision 1:
TV = Number of commercials for separate spots on television.
Decision 2: M = Number of advertisements in magazines.
Decision 3:
SS = Number of advertisements in Sunday supplements.
The decisions already made for the locations of the data cells have designated columns C,
D, and E to be the columns for these media. Therefore, the changing cells to hold these three
decisions have been placed in row 13 in these columns:
​​T V → cell C13​
M → cell D13​
SS → cell E13​​
These changing cells are collectively referred to by the range name NumberOfAds (C13:E13).
The Constraints
These changing cells need to be nonnegative. In addition, constraints are needed for the three
resources. The first two resources are the ad budget and planning budget. Data for these two
budgets already have ben placed in data cells in rows 8 and 9, so the constraints involving
these budgets need to appear in these same two rows. The amounts spent from these two
budgets are shown in the range BudgetSpent (F8:F9) and the amounts available from these
budgets are shown in the range of data cells BudgetAvailable (H8:H9). As suggested by the ≤
signs entered into column G, the corresponding constraints are
Total spending on advertising ≤ 4,000 (Ad budget in $1,000s)
​​
​
​
​
​​
Total cost of planning ≤ 1,000
(Planning budget in $1,000s)
Using the data in columns C, D, and E for the resources, these totals are
Total spending on advertising = 300TV + 150M + 100SS
​​
​
​  ​
​​​
Total cost of planning
= 90TV + 30M + 40SS
Excel Tip: Range
names may overlap. For
instance, we have used
­NumberOfAds to refer to
the whole range of changing cells, C13:E13, and
TVSpots to refer to the
single cell, C13.
These sums of products on the right-hand side are entered into the output cells BudgetSpent
(F8:F9) by using the SUMPRODUCT functions shown in the lower right-hand side of Figure
3.1. Although the ≤ signs entered in column G are only cosmetic (trial solutions still can be
entered in the changing cells that violate these inequalities), they will serve as a reminder later
to use these same ≤ signs when entering the constraints in Solver.
The third resource is TV spots for different commercials. Five such spots are available for
purchase. The number of spots used is one of the changing cells (C13). Since this cell will be
used in a constraint, we assign the cell its own range name: TVSpots (C13). The maximum
number of TV spots available is in the data cell MaxTVSpots (C15). Thus, the required constraint is TVSpots ≤ MaxTVSpots.
3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem 69
The Measure of Performance
Claire Syverson is using expected number of exposures as the overall measure of performance, so let
​Exposure = Expected number of exposures (in thousands) from all the advertising​
The data cells ExposuresPerAd (C4:E4) provide the expected number of exposures (in
­thousands) per advertisement in the respective media and the changing cells NumberOfAds
(C13:E13) give the number of each type of advertisement. Therefore,
Exposure = 1,300TV + 600M + 500SS
​​ ​
​  ​
​
​​
​
= SUMPRODUCT (ExposuresPerAd, NumberOfAds)
is the formula that needs to be entered into the objective cell, TotalExposures (H13).
Summary of the Formulation
The above analysis of the four components of the model has formulated the following linear
programming model (in algebraic form) on the spreadsheet:
​Maximize Exposure = 1,300TV + 600M + 500SS​
subject to
Ad spending:
300 TV + 150 M + 100 SS ≤ 4,000
costs:​
​​ Planning
​
​  90 TV ​  + 30 M + 40 SS​  ​  ≤​ ​  1,000​​​​
Number of television spots:
TV
≤
5
and
​​T V ≥ 0​
M ≥ 0​
SS ≥ 0​​
The difficult work of defining the problem and gathering all the relevant data, including the
information in Table 3.1, leads directly to this formulation.
Excel Tip: With Excel’s
Solver, the Solver dialog
box is used to tell Solver the
location on the spreadsheet
of several of the elements
of the model: the changing
cells, the objective cell, and
the constraints. With Analytic Solver, the Decisions,
Constraints, and Objective menu on the Analytic
Solver ribbon are used
along with the Model pane.
Solving the Model
To solve the spreadsheet model formulated above, some key information needs to be entered
into Solver. The lower left-hand side of Figure 3.1 shows the needed entries: the objective cell
(TotalExposures), the changing cells (NumberOfAds), the goal of maximizing the objective
cell, and the constraints BudgetSpent ≤ BudgetAvailable and TVSpots ≤ MaxTVSpots. Two
options are also specified at the bottom of the Solver Parameters box on the lower left-hand
side of Figure 3.1. The changing cells need nonnegativity constraints because negative values of
advertising are not possible. Choose the Simplex LP (Excel’s Solver) or Standard LP/Quadratic
Engine (Analytic Solver) solving method, because this is a linear programming model. Running
Solver then finds an optimal solution for the model and displays it in the changing cells.
The optimal solution given in row 13 of the spreadsheet provides the following plan for the
promotional campaign:
Do not run any television commercials.
Run 20 advertisements in magazines.
Run 10 advertisements in Sunday supplements.
Since TotalExposures (H13) gives the expected number of exposures in thousands, this plan
would be expected to provide 17,000,000 exposures.
Evaluation of the Adequacy of the Model
When she chose to use a linear programming model to represent this advertising-mix problem, Claire recognized that this kind of model does not provide a perfect match to this problem. However, a mathematical model is intended to be only an approximate representation
of the real problem. Approximations and simplifying assumptions generally are required to
have a workable model. All that is really needed is that there be a reasonably high correlation
between the prediction of the model and what would actually happen in the real problem. The
t eam now needs to check whether this criterion is satisfied.
70 Chapter Three Linear Programming: Formulation and Applications
Linear programming
models allow fractional
solutions.
Linear programming models should use SUM or
SUMPRODUCT functions
for the output cells, including the objective cell.
One assumption of linear programming is that fractional solutions are allowed. For the
current problem, this means that a fractional number (e.g., 3½) of television commercials
(or of ads in magazines or Sunday supplements) should be allowed. This is technically true,
since a commercial can be aired for less than a normal run, or an ad can be run in just a fraction of the usual magazines or Sunday supplements. However, one defect of the model is that
it assumes that Giacomi & Jackowitz’s cost for planning and developing a commercial or ad
that receives only a fraction of its usual run is only that fraction of its usual cost, even though
the actual cost would be the same as for a full run. Fortunately, the optimal solution obtained
was an integer solution (0 television commercials, 20 ads in magazines, and 10 ads in Sunday
supplements), so the assumption that fractional solutions are allowed was not even needed.
Although it is possible to have a fractional number of a normal run of commercials or
ads, a normal run tends to be much more effective than a fractional run. Therefore, it would
have been reasonable for Claire to drop the assumption that fractional solutions are allowed.
If Claire had done this and the optimal solution for the linear programming model had not
turned out to be integer, constraints can be added to require the changing cells to be integer.
(The TBA Airlines example in the next section provides an illustration of this type of constraint.) After adding such constraints, the model is called an integer programming model
instead of a linear programming model, but it still can be readily solved by Solver.
Another key assumption of linear programming is that the appropriate equation for each of
the output cells, including the objective cell, is one that can be expressed as a SUMPRODUCT
of data cells and changing cells (or occasionally just a SUM of changing cells). For the objective cell (cell H13) in Figure 3.1, this implies that the expected number of exposures to be
obtained from each advertising medium is proportional to the number of advertisements in
that medium. This proportionality seems true, since each viewing of the advertisements by
some individual counts as another exposure. Another implication of using a SUMPRODUCT
function is that the expected number of exposures to be obtained from an advertising medium
is unaffected by the number of advertisements in the other media. Again, this implication
seems valid, since viewings of advertisements in different media count as separate exposures.
Although a SUMPRODUCT function is appropriate for calculating the expected number
of exposures, the choice of this number for the overall measure of performance is somewhat
questionable. Management’s real objective is to maximize the profit generated as a result of
the advertising campaign, but this is difficult to measure so expected number of exposures
was selected to be a surrogate for profit. This would be valid if profit were proportional to the
expected number of exposures. However, proportionality is only an approximation in this case
because too many exposures for the same individual reach a saturation level where the impact
(potential profit) from one more exposure is substantially less than for the first exposure.
(When proportionality is not a reasonable approximation, Chapter 8 will describe nonlinear
models that can be used instead.)
To check how reasonable it is to use expected number of exposures as a surrogate for
profit, Claire meets with Sid Jackowitz, one of the senior partners of Giacomi & Jackowitz.
Sid indicates that the contemplated promotional campaign (20 advertisements in magazines
and 10 in Sunday supplements) is a relatively modest one well below saturation levels. Most
readers will only notice these ads once or twice, and a second notice is very helpful for reinforcing the first one. Furthermore, the readership of magazines and Sunday supplements is
sufficiently different that the interaction of the advertising impact in these two media should
be small. Consequently, Claire concludes that using expected number of exposures for the
objective cell in Figure 3.1 provides a reasonable approximation. (A continuation of this case
study in Case 8-1 will delve into the more complicated analysis that is required in order to use
profit directly as the measure of performance to be recorded in the objective cell instead of
making this approximation.)
Next, Claire quizzes Sid about his firm’s costs for planning and developing advertisements
in these media. Is it reasonable to assume that the cost in a given medium is proportional
to the number of advertisements in that medium? Is it reasonable to assume that the cost of
developing advertisements in one medium would not be substantially reduced if the firm had
just finished developing advertisements in another medium that might have similar themes?
Sid acknowledges that there is some carryover in ad planning from one medium to another,
3.2 Resource-Allocation Problems 71
especially if both are print media (e.g., magazines and Sunday supplements), but that the
carryover is quite limited because of the distinct differences in these media. Furthermore,
he feels that the proportionality assumption is quite reasonable for any given medium since
the amount of work involved in planning and developing each additional advertisement in
the medium is nearly the same as for the first one in the medium. The total fee that Super
Grain will pay Giacomi & Jackowitz will eventually be based on a detailed accounting of the
amount of work done by the firm. Nevertheless, Sid feels that the cost estimates previously
provided by the firm (as entered in cells C9, D9, and E9 in units of thousands of dollars) give
a reasonable basis for roughly projecting what the fee will be for any given plan (the entries in
the changing cells) for the promotional campaign.
Based on this information, Claire concludes that using a SUMPRODUCT function for cell
F9 provides a reasonable approximation. Doing the same for cell F8 is clearly justified. Given
her earlier conclusions as well, Claire decides that the linear programming model incorporated into Figure 3.1 (plus any expansions of the model needed later for the detailed planning)
is a sufficiently accurate representation of the real advertising-mix problem. It will not be
necessary to refine the results from this model by turning next to a more complicated kind of
mathematical model (such as those to be described in Chapter 8).
Therefore, Claire sends a memorandum to the company’s president, David Sloan, describing a promotional campaign that corresponds to the optimal solution from the linear programming model (no TV commercials, 20 ads in magazines, and 10 ads in Sunday supplements).
She also requests a meeting to evaluate this plan and discuss whether some modifications
should be made.
We will pick up this story again in Section 3.4.
Review
Questions
3.2
1. What is the problem being addressed in this case study?
2. What overall measure of performance is being used?
3. What are the assumptions of linear programming that need to be checked to evaluate the adequacy of using a linear programming model to represent the problem under consideration?
RESOURCE-ALLOCATION PROBLEMS
In the opening paragraph of Chapter 2, we described managerial problems involving the allocation of an organization’s resources to its various productive activities. Those were resourceallocation problems.
Resource-allocation problems are linear programming problems involving the allocation of
resources to activities. The identifying feature for any such problem is that each functional constraint in the linear programming model is a resource constraint, which has the form
​Amount of resource used ≤ Amount of resource available​
for one of the resources.
The amount of a resource used depends on which activities are undertaken, the levels of
those activities, and how heavily those activities need to use the resource. Thus, the resource
constraints place limits on the levels of the activities. The objective is to choose the levels of
the activities so as to maximize some overall measure of performance (such as total profit)
from the activities while satisfying all the resource constraints.
Beginning with the Super Grain case study and then the Wyndor case study from ­Chapter 2,
we will look at four examples that illustrate the characteristics of resource-allocation problems. These examples also demonstrate how this type of problem can arise in a variety of
contexts.
The Super Grain Corp. Advertising-Mix Problem
The linear programming model formulated in Section 3.1 for the Super Grain case study is
one example of a resource-allocation problem. The three activities under consideration are the
advertising in the three types of media chosen by Giacomi & Jackowitz.
72 Chapter Three Linear Programming: Formulation and Applications
Activity 1: TV commercials
Activity 2: Magazine ads
Activity 3: Sunday ads
An initial step in formulating any resource-allocation
problem is to identify the
activities and the resources.
The decisions being made are the levels of these activities, that is, the number of TV commercials, magazine ads, and Sunday ads to run.
The resources to be allocated to these activities are
Resource 1: Advertising budget ($4 million).
Resource 2: Planning budget ($1 million).
Resource 3: TV spots available for different commercials (5).
where the amounts available of these resources are given in parentheses. Thus, this problem
has three resource constraints:
1. Advertising budget used
2. Planning budget used
3. TV spots used
≤ $4 million
≤ $1 million
≤5
Rows 8–9 and cells C13:C15 in Figure 3.1 show these constraints in a spreadsheet. Cells
C8:E9 give the amount of the advertising budget and the planning budget used by each unit
of each activity, that is, the amount used by one TV spot, one magazine ad, and one Sunday
ad, respectively.
Cells C4, D4, and E4 on this spreadsheet give the contribution per unit of each activity to
the overall measure of performance (expected number of exposures).
Characteristics of Resource-Allocation Problems
For each proposed activity, a decision needs to be
made as to how much of
the activity to do. In other
words, what should the
level of the activity be?
These three kinds of data
are needed for any resourceallocation problem.
Other resource-allocation problems have the same kinds of characteristics as the Super Grain
problem. In each case, there are activities where the decisions to be made are the levels of
these activities. The contribution of each activity to the overall measure of performance is
proportional to the level of that activity. Commonly, this measure of performance is the total
profit from the activities, but occasionally it is something else (as in the Super Grain problem).
Every problem of this type has a resource constraint for each resource. The amounts of the
resources used depend on the levels of the activities. For each resource, the amount used by
each activity is proportional to the level of that activity.
The manager or management science team studying a resource allocation problem needs to
gather (with considerable help) three kinds of data:
1. The amount available of each resource.
2. The amount of each resource needed by each activity. Specifically, for each combination of
resource and activity, the amount of the resource used per unit of the activity must be estimated.
3. The contribution per unit of each activity to the overall measure of performance.
Generally there is considerable work involved in developing these data. A substantial
amount of digging and consultation is needed to obtain the best estimates available in a timely
fashion. This step is critical. Well-informed estimates are needed to obtain a valid linear programming model for guiding managerial decisions. The dangers involved in inaccurate estimates are one reason why what-if analysis (Chapter 5) is such an important part of most linear
programming studies.
The Wyndor Glass Co. Product-Mix Problem
The product-mix problem facing the management of the Wyndor Glass Co. in Section 2.1 is
to determine the most profitable mix of production rates for the two new products, considering the limited availability of spare production capacity in the company’s three plants. This is
a resource-allocation problem.
The activities under consideration are
Activity 1: Produce the special new doors.
Activity 2: Produce the special new windows.
3.2 Resource-Allocation Problems 73
The decisions being made are the levels of these activities, that is, the production rates for the
doors and windows. Production rate is being measured as the number of units (doors or windows) produced per week. Management’s objective is to maximize the total profit generated
by the two new products, so the overall measure of performance is total profit. The contribution of each product to profit is proportional to the production rate for that product.
The resources to be allocated to these activities are
Resource 1: Production capacity in Plant 1.
Resource 2: Production capacity in Plant 2.
Resource 3: Production capacity in Plant 3.
Each of the three functional constraints in the linear programming model formulated in
Section 2.2 (see rows 7–9 of the spreadsheet in Figure 2.3 or 2.4) is a resource constraint for
one of these three resources. Column E shows the amount of production capacity used in each
plant and column G gives the amount available.
Table 2.1 in Section 2.1 provides the data for the Wyndor problem. You already have seen
how the numbers in Table 2.1 become the parameters in the linear programming model in
either its spreadsheet formulation (Section 2.2) or its algebraic form (Section 2.3).
The TBA Airlines Problem
TBA Airlines is a small regional company that specializes in short flights in small passenger airplanes. The company has been doing well and management has decided to expand its operations.
The Problem
The basic issue facing management now is whether to purchase more small airplanes to add
some new short flights or to start moving into the national market by purchasing some large
airplanes for new cross-country flights (or both). Many factors will go into management’s
final decision, but the most important one is which strategy is likely to be most profitable.
The first row of Table 3.2 shows the estimated net annual profit (inclusive of capital recovery costs) from each type of airplane purchased. The second row gives the purchase cost per
airplane and also notes that the total amount of capital available for airplane purchases is
$250 million. The third row records the fact that management does not want to purchase more
than five small airplanes because of limited possibilities for adding lucrative short flights,
whereas they have not specified a maximum number for large airplanes (other than that
imposed by the limited capital available).
How many airplanes of each type should be purchased to maximize the total net annual profit?
Formulation
This is a resource-allocation problem. The activities under consideration are
Activity 1: Purchase small airplanes.
Activity 2: Purchase large airplanes.
The decisions to be made are the levels of these activities, that is,
S = Number of small airplanes to purchase
L = Number of large airplanes to purchase
The one resource to be allocated to these activities is
Resource: Investment capital ($250 million).
Thus, there is a single resource constraint:
​Investment capital spent ≤ $250 million​
TABLE 3.2
Data for the TBA
Airlines Problem
Small Airplane
Net annual profit per airplane
Purchase cost per airplane
Maximum purchase quantity
$7 million
$25 million
5
Large Airplane
$22 million
$75 million
No maximum
Capital Available
$250 million
74 Chapter Three Linear Programming: Formulation and Applications
FIGURE 3.2
A spreadsheet model for the TBA Airlines integer programming problem where the changing cells, UnitsProduced (C12:D12), show the
optimal airplane purchases obtained by Solver, and the objective cell, TotalProfit (G12), gives the resulting total profit in millions of dollars.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
B
C
D
E
F
G
<=
Capital
Available
250
TBA Airlines Airplane Purchasing Problem
Unit Profit ($millions)
Small Airplane
7
Capital per Unit Purchased
25
75
Capital ($millions)
Number Purchased
Maximum Small Airplanes
6
7
8
Small Airplane
1
<=
5
Capital
Spent
250
Large Airplane
3
Total Profit
($millions)
73
E
Capital
Spent
=SUMPRODUCT(CapitalPerUnitPurchased,NumberPurchased)
Solver Parameters
Set Objective Cell: TotalProfit
To: Max
By Changing Variable Cells:
NumberPurchased
Subject to the Constraints:
CapitalSpent <= CapitalAvailable
NumberPurchased = integer
SmallAirplanes <= MaxSmallAirplanes
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
Large Airplane
22
10
11
12
G
Total Profit
($millions)
=SUMPRODUCT(UnitProfit,NumberPurchased)
Range Name
Capital Available
CapitalPerUnitPurchased
CapitalSpent
MaxSmallAirplanes
NumberPurchased
SmallAirplanes
TotalProfit
UnitProfit
Cells
G8
C8:D8
E8
C14
C12:D12
C12
G12
C4:D4
In addition, management has specified one side constraint:
​Number of small airplanes purchased ≤ 5​
Figure 3.2 shows the formulation of a spreadsheet model for this problem, where the data
in Table 3.2 have been transferred into the data cells—UnitProfit (C4:D4), CapitalPerUnitPurchased (C8:D8), CapitalAvailable (G8), and MaxSmallAirplanes (C14). The resource constraint then appears in cells C8:G8 while C12:C14 shows the side constraint. The objective for
this problem is to maximize the total net annual profit, so the equation for the objective cell is
​TotalProfit ​​(​​G12​)​​​ = SUMPRODUCT (​​ ​​UnitProfit, UnitsPurchased​)​​​​
Since the TBA Airlines problem is a resource-allocation problem, this spreadsheet model
has essentially the same form as the Super Grain and Wyndor problems except for one small
difference. The changing cells in this case must have integer values since it is not feasible for
the company to purchase and operate a fraction of an airplane. Therefore, constraints that the
changing cells need to be integer are added. With Excel’s Solver, use the Add Constraint dialog box to choose the range of these cells (C12:D12) as the left-hand side and then choose int
3.2 Resource-Allocation Problems 75
Excel Tip: To constrain a
range of changing cells to
be integer in Excel’s Solver,
choose the range of cells
in the left-hand side of the
Add Constraint dialog box
and choose int from the
pop-up menu. Clicking OK
then enters the constraint
that these cells = integer in
the Solver dialog box. In
Analytic Solver, select the
range of cells to be constrained integer, and then
under the Constraint menu
on the Analytic Solver ribbon, choose Integer under
the Variable Type/Bound
submenu.
Excel Tip: Even when a
changing cell is constrained
to be integer, rounding
errors occasionally will
cause Excel to return a noninteger value very close to
an integer (e.g., 1.23E-10,
meaning 0.000000000123).
To make the spreadsheet
cleaner, you may replace
these “ugly” representations
by their proper integer values in the changing cells.
Excel Tip: In Excel’s Solver
Options, the Integer Optimality (%) setting
(1 percent by default) causes
Solver to stop solving an
integer programming problem when it finds a feasible
solution whose objective
function value is within
the specified percentage of
being optimal. In Analytic
Solver, the equivalent setting
is Integer Tolerance under
the Engine tab of the Solver
model (see Figure 2.19).
This is useful to speed up
solving large problems.
For smaller problems (e.g.,
homework problems), this
option should be set to 0 to
guarantee finding an optimal
solution.
from the pop-up menu between the left-hand and right-hand side. In Analytic Solver, choose
the changing cells (C12:D12), and then under the Constraint menu on the Analytic Solver ribbon, choose Integer under the Variable Type/Bound submenu.
These changing cells in Figure 3.2 show the optimal solution, (S, L) = (1, 3), obtained after
running Solver.
One of the assumptions of linear programming is that the changing cells are allowed to
have any values, including fractional values, that satisfy the functional and nonnegativity
constraints. Therefore, technically speaking, the TBA problem is not a linear programming
problem because of adding the constraints
​NumberPurchased = integer​
that are displayed in the Solver Parameters box in Figure 3.2. Such a problem that fits linear
programming except for adding such constraints is called an integer programming problem.
The method used by Solver to solve integer programming problems is quite different from
that for solving linear programming problems. In fact, integer programming problems tend
to be much more difficult to solve than linear programming problems so there is considerably more limitation on the size of the problem. However, this doesn’t matter to a spreadsheet
modeler dealing with small problems. From his or her viewpoint, there is virtually no distinction between linear programming and integer programming problems. They are formulated in
exactly the same way. Then, at the very end, a decision needs to be made as to whether any of
the changing cells need to be restricted to integer values. If so, those constraints are added as
described above. Keep this option in mind as we continue to discuss the formulation of various types of linear programming problems throughout the chapter.
Summary of the Formulation
The above formulation of a model with one resource constraint and one side constraint for the
TBA Airlines problem now can be summarized (in algebraic form) as follows:
​​Maximize
subject to
and
​  Profit = 7S + 22L​​
25S + 75L ≤ 250
​​
​  ​  ​
​​​
S
≤ 5
​​S ≥ 0​
L ≥ 0​​
Capital Budgeting
Financial planning is one of the most important areas of application for resource-allocation
problems. The resources being allocated in this area are quite different from those for applications in the production planning area (such as the Wyndor Glass Co. product-mix problem),
where the resources tend to be production facilities of various kinds. For financial planning,
the resources tend to be financial assets such as cash, securities, accounts receivable, lines of
credit, and so forth. Our specific example involves capital budgeting, where the resources are
amounts of investment capital available at different points in time.
The Problem
The Think-Big Development Co. is a major investor in commercial real-estate development
projects. It currently has the opportunity to share in three large construction projects:
Project 1: Construct a high-rise office building.
Project 2: Construct a hotel.
Project 3: Construct a shopping center.
Each project requires each partner to make investments at four different points in time: a
down payment now, and additional capital after one, two, and three years. Table 3.3 shows for
each project the total amount of investment capital required from all the partners at these four
points in time. Thus, a partner taking a certain percentage share of a project is obligated to
invest that percentage of each of the amounts shown in the table for the project.
An Application Vignette
A key part of a country’s financial infrastructure is its securities
markets. By allowing a variety of financial institutions and their
clients to trade stocks, bonds, and other financial securities,
they help fund both public and private initiatives. Therefore, the
efficient operation of its securities markets plays a crucial role
in providing a platform for the economic growth of the country.
Each central securities depository and its system for quickly
settling security transactions are part of the operational backbone of securities markets and a key component of financial
system stability. In Mexico, an institution called INDEVAL provides both the central securities depository and its security
settlement system for the entire country. This security settlement system uses electronic book entries, modifying cash and
securities balances, for the various parties in the transactions.
The total value of the securities transactions the INDEVAL
settles averages over $250 billion daily. This makes INDEVAL
the main liquidity conduit for Mexico’s entire financial sector.
Therefore, it is extremely important that INDEVAL’s system for
clearing securities transactions be an exceptionally efficient one
that maximizes the amount of cash that can be delivered almost
instantaneously after the transactions. Because of past dissatisfaction with this system, INDEVAL’s board of directors ordered a
major study in 2005 to completely redesign the system.
Following more than 12,000 man-hours devoted to this redesign, the new system was successfully launched in N
­ ovember
2008. The core of the new system is a large linear programming model that is applied many times daily to choose which
pending transactions should be settled immediately with the
depositor’s available balances. Linear programming is ideally
suited for this application because it can maximize the value of
the transactions settled while taking into account the various
relevant constraints.
This application of linear programming has substantially
enhanced and strengthened the Mexican financial infrastructure by reducing its daily liquidity requirements by $130 billion.
It also reduces the intraday financing costs for market participants by more than $150 million annually. This application
also led to INDEVAL winning the prestigious first prize in the
2010 international competition for the Franz Edelman Award
for Achievement in Operations Research and the Management
Sciences.
Source: D. Muñoz, M. de Lascurain, O. Romeo-Hernandez, F. Solis, L.
de los Santoz, A. Palacios-Brun, F. Herrería, and J. Villaseñor, “INDEVAL
Develops a New Operating and Settlement System Using Operations
Research,” Interfaces 41, no. 1 (January–February 2011), pp. 8–17. (A link
to this article is available at www.mhhe.com/Hillier6e.)
TABLE 3.3
Financial Data for the
Projects Being Considered for Partial Investment by the Think-Big
Development Co.
Investment Capital Requirements
Year
Office Building
Hotel
Shopping Center
0
1
2
3
$40 million
60 million
90 million
10 million
$80 million
80 million
80 million
70 million
$90 million
50 million
20 million
60 million
$45 million
$70 million
$50 million
Net present value
76
All three projects are expected to be very profitable in the long run. So the management of
Think-Big wants to invest as much as possible in some or all of them. Management is willing
to commit all the company’s investment capital currently available, as well as all additional
investment capital expected to become available over the next three years. The objective is
to determine the investment mix that will be most profitable, based on current estimates of
profitability.
Since it will be several years before each project begins to generate income, which will
continue for many years thereafter, we need to take into account the time value of money
in evaluating how profitable it might be. This is done by discounting future cash outflows
(capital invested) and cash inflows (income), and then adding discounted net cash flows, to
calculate a project’s net present value.
Based on current estimates of future cash flows (not included here except for outflows),
the estimated net present value for each project is shown in the bottom row of Table 3.3. All
the investors, including Think-Big, then will split this net present value in proportion to their
share of the total investment.
For each project, participation shares are being sold to major investors, such as ThinkBig, who become the partners for the project by investing their proportional shares at the four
specified points in time. For example, if Think-Big takes a 10 percent share of the office building, it will need to provide $4 million now, and then $6 million, $9 million, and $1 ­million in
1 year, 2 years, and 3 years, respectively.
3.2 Resource-Allocation Problems 77
The company currently has $25 million available for capital investment. Projections are
that another $20 million will become available after one year, $20 million more after two
years, and another $15 million after three years. What share should Think-Big take in the
respective projects to maximize the total net present value of these investments?
Formulation
This is a resource-allocation problem. The activities under consideration are
Activity 1: Invest in the construction of an office building.
Activity 2: Invest in the construction of a hotel.
Activity 3: Invest in the construction of a shopping center.
Thus, the decisions to be made are the levels of these activities, that is, what participation
share to take in investing in each of these projects. A participation share can be expressed as
either a fraction or a percentage of the entire project, so the entire project is considered to be
one “unit” of that activity.
The resources to be allocated to these activities are the funds available at the four investment points. Funds not used at one point are available at the next point. (For simplicity, we
will ignore any interest earned on these funds.) Therefore, the resource constraint for each
point must reflect the cumulative funds to that point.
Resource 1: Total investment capital available now.
Resource 2: Cumulative investment capital available by the end of one year.
Resource 3: Cumulative investment capital available by the end of two years.
Resource 4: Cumulative investment capital available by the end of three years.
Since the amount of investment capital available is $25 million now, another $20 million
in one year, another $20 million in two years, and another $15 million in three years, the
amounts available of the resources are the following:
Amount of resource 1 available = $25million
Amount of resource 2 available = $(25 + 20) million = $45 million
Amount of resource 3 available = $(25 + 20 + 20) million = $65 million
Amount of resource 4 available = $(25 + 20 + 20 + 15) million = $80 million
Table 3.4 shows all the data involving these resources. The rightmost column gives the
amounts of resources available calculated above. The middle columns show the cumulative
amounts of the investment capital requirements listed in Table 3.3. For example, in the Office
Building column of Table 3.4, the second number ($100 million) is obtained by adding the
first two numbers ($40 million and $60 million) in the Office Building column of Table 3.3.
The Data As with any resource-allocation problem, three kinds of data need to be gathered.
One is the amounts available of the resources, as given in the rightmost column of Table 3.4.
A second is the amount of each resource needed by each project, which is given in the middle
columns of this table. A third is the contribution of each project to the overall measure of
performance (net present value), as given in the bottom row of Table 3.3.
The first step in formulating the spreadsheet model is to enter these data into data cells
in the spreadsheet. In Figure 3.3, the data cells (and their range names) are NetPresentValue
TABLE 3.4
Resource Data for the
Think-Big Development
Co. Investment-Mix
Problem
Cumulative Investment Capital Required for an Entire Project
Resource
1 (Now)
2 (End of year 1)
3 (End of year 2)
4 (End of year 3)
Office
Building
Hotel
Shopping
Center
Amount of
Resource Available
$ 40 million
100 million
190 million
200 million
$ 80 million
160 million
240 million
310 million
$ 90 million
140 million
160 million
220 million
$25 million
45 million
65 million
80 million
78 Chapter Three Linear Programming: Formulation and Applications
FIGURE 3.3
The spreadsheet model for the Think-Big problem, including the formulas for the objective cell TotalNPV (H16) and the other output
cells CapitalSpent (F9:F12), as well as the specifications needed to set up Solver. The changing cells ParticipationShare (C16:E16)
show the optimal solution obtained by Solver.
A
1
B
C
D
E
F
G
H
Think-Big Development Co. Capital Budgeting Program
2
3
Office
4
Building
Hotel
Center
45
70
50
5
Net Present Value
6
($millions)
Shopping
Cumulative
7
Cumulative Capital Required ($millions)
8
Cumulative
Capital
Capital
Spent
Available
9
Now
40
80
90
25
≤
25
10
End of Year 1
100
160
140
44.76
≤
45
11
End of Year 2
190
240
160
60.58
≤
65
12
End of Year 3
200
310
220
80
≤
80
13
14
Office
Shopping
Total NPV
15
Building
Hotel
Center
($millions)
0.00%
16.50%
13.11%
18.11
16
Participation Share
Solver Parameters
Set Objective Cell: TotalNPV
To: Max
By Changing Variable Cells:
ParticipationShare
Subject to the Constraints:
CapitalSpent <= CapitalAvailable
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
F
Range Name
Cells
CapitalAvailable
CapitalRequired
CapitalSpent
ParticipationShare
NetPresentValue
TotalNPV
H9:H12
C9:E12
F9:F12
C16:E16
C5:E5
H16
6
Cumulative
7
Capital
8
9
10
=SUMPRODUCT(C10:E10,ParticipationShare)
11
12
=SUMPRODUCT(C11:E11,ParticipationShare)
=SUMPRODUCT C12:E12,ParticipationShare)
Spent
=SUMPRODUCT(C9:E9,ParticipationShare)
H
14
Total NPV
15
($millions)
16
=SUMPRODUCT(NetPresentValue,ParticipationShare)
(C5:E5), CapitalRequired (C9:E12), and CapitalAvailable (H9:H12). To save space on the
spreadsheet, these numbers are entered in units of millions of dollars.
The Decisions With three activities under consideration, there are three decisions to be made.
Decision 1:
Decision 2:
Decision 3:
OB = Participation share in the office building.
H = Participation share in the hotel.
SC = Participation share in the shopping center.
For example, if Think-Big management were to decide to take a one-tenth participation share
(i.e., a 10 percent participation share) in each of these projects, then
OB = 0.1 = 10%
H = 0.1 = 10%
SC = 0.1 = 10%
However, it may not be desirable to take the same participation share (expressed as either a
fraction or a percentage) in each of the projects, so the idea is to choose the best combination
3.2 Resource-Allocation Problems 79
of values of OB, H, and SC. In Figure 3.3, the participation shares (expressed as percentages)
have been placed in changing cells under the data cells (row 16) in the columns for the three
projects, so
​​OB → cell C16​
H → D16​
SC → cell E16​​
where these cells are collectively referred to by the range name ParticipationShare (C16:E16).
The Constraints The numbers in these changing cells make sense only if they are nonnegative, so the Make Variables Nonnegative option will need to be selected in the Excel’s
Solver dialog box (or equivalently in Analytic Solver, set the Assume Non-Negative option
to True in the Engine tab of the Model pane). In addition, the four resources require resource
constraints:
Total invested now
≤ 25 (millions of dollars available)
Total invested within 1 year ≤ 45 (millions of dollars available)
Total invested within 2 years ≤ 65 (millions of dollars available)
Total invested within 3 years ≤ 80 (millions of dollars available)
The data in columns C, D, and E indicate that (in millions of dollars)
Total invested now
= 40 OB + 80 H + 90 SC
Total invested within 1 year = 100 OB + 160 H + 140 SC
Total invested within 2 years = 190 OB + 240 H + 160 SC
Total invested within 3 years = 200 OB + 310 H + 220 SC
These totals are calculated in the output cells CapitalSpent (F9:F12) using the SUMPRODUCT
function, as shown below the spreadsheet in Figure 3.3. Finally, ≤ signs are entered into column G to indicate the resource constraints that will need to be entered in Solver.
The Measure of Performance The objective is to
​​Maximize
​  NPV = total net present value of the investments​​
NetPresentValue (C5:E5) shows the net present value of each entire project, while
­ParticipationShare (C16:E16) shows the participation share for each of the projects. Therefore, the total net present value of all the participation shares purchased in all three projects is
(in millions of dollars)
NPV = 45 OB + 70 H + 50 SC
ParticipationShare​)​​​​​​
​ ​​​
​  =​ ​  ​  SUMPRODUCT (​​ ​​NetPresentValue,
​
​
→ cell H16
Summary of the Formulation This completes the formulation of the linear programming
model on the spreadsheet, as summarized below (in algebraic form).
​​Maximize
​  NPV = 45 OB + 70 H + 50 SC​​
subject to
Total invested now:
40 OB + 80 H + 90 SC ≤ 25
Total invested within 1 year: 100 OB + 160 H + 140 SC ≤ 45
Total invested within 2 years: 190 OB + 240 H + 160 SC ≤ 65
Total invested within 3 years: 200 OB + 310 H + 220 SC ≤ 80
and
​​OB ≥ 0​
H ≥ 0​
SC ≥ 0​​
where all these numbers are in units of millions of dollars.
80 Chapter Three Linear Programming: Formulation and Applications
Note that this model possesses the key identifying feature for resource-allocation problems, namely, each functional constraint is a resource constraint that has the form
​Amount of resource used ≤ Amount of resource available​
Solving the Model The lower left-hand side of Figure 3.3 shows the entries needed in
Solver to specify the model, along with the selection of the usual two options. The spreadsheet shows the resulting optimal solution in row 16, namely,
Invest nothing in the office building.
Invest in 16.50 percent of the hotel.
Invest in 13.11 percent of the shopping center.
TotalNPV (H16) indicates that this investment program would provide a total net present
value of $18.11 million.
This amount actually is only an estimate of what the total net present value would turn out
to be, depending on the accuracy of the financial data given in Table 3.3. There is some uncertainty about the construction costs for the three real estate projects, so the actual investment
capital requirements for years 1, 2, and 3 may deviate somewhat from the amounts specified
in this table. Because of the risk involved in these projects, the net present value for each one
also might deviate from the amounts given at the bottom of the table. Chapter 5 describes
one approach to analyzing the effect of such deviations. Chapters 12 and 13 will present
another technique, called computer simulation, for systematically taking future uncertainties
into account. Section 13.4 will focus on further analysis of this same example.
Another Look at Resource Constraints
These examples of resource-allocation problems illustrate a variety of resources: financial
allocations for advertising and planning purposes, TV commercial spots available for purchase, available production capacities of different plants, the total amount of capital available for investment, and cumulative investment capital available by certain times. However,
these illustrations only scratch the surface of the realm of possible resources that need to be
allocated to activities in resource-allocation problems. In fact, by interpreting resource sufficiently broadly, any restriction on the decisions to be made that has the form
​Amount used ≤ Amount available​
can be thought of as a resource constraint, where the thing whose amount is being measured
is the corresponding “resource.” Since any functional constraint with a ≤ sign in a linear
programming model (including the side constraint limiting the number of small airplanes
that can be purchased in the TBA Airlines example) can be verbalized in this form, any such
constraint can be thought of as a resource constraint.
Hereafter, we will use resource constraint to refer to any functional constraint with a
≤ sign in a linear programming model. The constant on the right-hand side represents the
amount available of a resource. Therefore, the left-hand side represents the amount used of this
resource. In the algebraic form of the constraint, the coefficient (positive or negative) of each
decision variable is the resource usage per unit of the corresponding activity.
Summary of the Formulation Procedure for
Resource-Allocation Problems
The four examples illustrate that the following steps are used for any resource-allocation problem to define the specific problem, gather the relevant data, and then formulate the linear
programming model.
1. Since any linear programming problem involves finding the best mix of levels of various
activities, identify these activities for the problem at hand. The decisions to be made are
the levels of these activities.
2. From the viewpoint of management, identify an appropriate overall measure of performance (commonly profit, or a surrogate for profit) for solutions of the problem.
3.3 Cost–Benefit–Trade-Off Problems 81
FIGURE 3.4
A template of a spreadsheet model for pure resource-allocation problems.
Activities
Constraints
Unit Profit
Profit per unit of activity
Resource used per unit of activity
Resources
Used
SUMPRODUCT
(resource used per unit,
changing cells)
Resources
Available
≤
Total Profit
Level of Activity
Changing cells
SUMPRODUCT(profit per unit, changing cells)
3. For each activity, estimate the contribution per unit of the activity to this overall measure
of performance.
4. Identify the resources that must be allocated to the activities.
5. For each resource, identify the amount available and then the amount used per unit of each
activity.
6. Enter the data gathered in steps 3 and 5 into data cells in a spreadsheet. A convenient format is to have the data associated with each activity in a separate column, the data for the
unit profit and each constraint in a separate row, and to leave two blank columns between
the activity columns and the amount of resource available column. Figure 3.4 shows a
template of the overall format of a spreadsheet model for resource-allocation problems.
7. Designate changing cells for displaying the decisions on activity levels.
8. For the two blank columns created in step 6, use the left one as a Totals column for output
cells and enter ≤ signs into the right one for all the resources. In the row for each resource,
use the SUMPRODUCT function to enter the total amount used in the Totals column.
9. Designate an objective cell for displaying the overall measure of performance. Use a
SUMPRODUCT function to enter this measure of performance.
All the functional constraints in this linear programming model in a spreadsheet are
resource constraints, that is, constraints with a ≤ sign. This is the identifying feature that classifies the problem as being a resource-allocation problem.
Review
Questions
3.3
1.
2.
3.
4.
5.
What is the identifying feature for a resource-allocation problem?
What is the form of a resource constraint?
What are the three kinds of data that need to be gathered for a resource-allocation problem?
Compare the types of activities for the four examples of resource-allocation problems.
Compare the types of resources for the four examples of resource-allocation problems.
COST–BENEFIT–TRADE-OFF PROBLEMS
Cost–benefit–trade-off problems have a form that is very different from resource-allocation
problems. The difference arises from managerial objectives that are very different for the two
kinds of problems.
For resource-allocation problems, limits are set on the use of various resources (including
financial resources), and then the objective is to make the most effective use (according to
some overall measure of performance) of these given resources.
For cost–benefit–trade-off problems, management takes a more aggressive stance, prescribing what benefits must be achieved by the activities under consideration (regardless of
82 Chapter Three Linear Programming: Formulation and Applications
A cost–benefit–trade-off
formulation enables management to specify minimum goals for the benefits
that need to be achieved by
the activities.
the resulting resource usage), and then the objective is to achieve all these benefits with minimum cost. By prescribing a minimum acceptable level for each kind of benefit, and then minimizing the cost needed to achieve these levels, management hopes to obtain an appropriate
trade-off between cost and benefits. (You will see in Chapter 5 that what-if analysis plays a
key role in providing the additional information needed for management to choose the best
trade-off between cost and benefits.)
Cost–benefit–trade-off problems are linear programming problems where the mix of levels
of various activities is chosen to achieve minimum acceptable levels for various benefits at a
minimum cost. The identifying feature is that each functional constraint is a benefit constraint,
which has the form
​Level achieved ≥ Minimum acceptable level​
for one of the benefits.
These three kinds of data
are needed for any cost–
benefit–trade-off problem.
Interpreting benefit broadly, we can think of any functional constraint with a ≥ sign as
a benefit constraint. In most cases, the minimum acceptable level will be prescribed by
management as a policy decision, but occasionally this number will be dictated by other
circumstances.
For any cost–benefit–trade-off problem, a major part of the study involves identifying all
the activities and benefits that should be considered and then gathering the data relevant to
these activities and benefits.
Three kinds of data are needed:
1. The minimum acceptable level for each benefit (a managerial policy decision).
2. For each benefit, the contribution of each activity to that benefit (per unit of the activity).
3. The cost per unit of each activity.
Let’s examine two examples of cost–benefit–trade-off problems.
The Profit & Gambit Co. Advertising-Mix Problem
An initial step in formulating any cost–benefit–­
tradeoff problem is to
identify the activities and
the benefits.
As described in Section 2.7, the Profit & Gambit Co. will be undertaking a major new advertising campaign focusing on three cleaning products. The two kinds of advertising to be used
are television and the print media. Management has established minimum goals—the minimum acceptable increase in sales for each product—to be gained by the campaign.
The problem is to determine how much to advertise in each medium to meet all the sales
goals at a minimum total cost.
The activities in this cost–benefit–trade-off problem are
Activity 1: Advertise on television.
Activity 2: Advertise in the print media.
The benefits being sought from these activities are
Benefit 1: Increased sales for a spray prewash stain remover.
Benefit 2: Increased sales for a liquid laundry detergent.
Benefit 3: Increased sales for a powder laundry detergent.
Management wants these increased sales to be at least 3 percent, 18 percent, and 4 percent,
respectively. As shown in Section 2.7, each benefit leads to a benefit constraint that incorporates the managerial goal for the minimum acceptable level of increase in the sales for the
corresponding product, namely,
Level of benefit 1 achieved ≥ 3%
Level of benefit 2 achieved ≥ 18%
Level of benefit 3 achieved ≥ 4%
The data for this problem are given in Table 2.2 (Section 2.7). Section 2.7 describes how
the linear programming model is formulated directly from the numbers in this table.
This example provides an interesting contrast with the Super Grain Corp. case study in ­Section
3.1, which led to a formulation as a resource-allocation problem. Both are advertising-mix
3.3 Cost–Benefit–Trade-Off Problems 83
problems, yet they lead to entirely different linear programming models. They differ because of
the differences in the managerial view of the key issues in each case:
• As the vice president for marketing of Super Grain, Claire Syverson focused first on how
much to spend on the advertising campaign and then set limits (an advertising budget of
$4 million and a planning budget of $1 million) that led to resource constraints.
• The management of Profit & Gambit instead focused on what it wanted the advertising
campaign to accomplish and then set goals (minimum required increases in sales) that led
to benefit constraints.
From this comparison, we see that it is not the nature of the application that determines the
classification of the resulting linear programming formulation. Rather, it is the nature of the
restrictions imposed on the decisions regarding the mix of activity levels. If the restrictions
involve limits on the usage of resources, that identifies a resource-allocation problem. If the
restrictions involve goals on the levels of benefits, that characterizes a cost–benefit–trade-off
problem. Frequently, the nature of the restrictions arise from the way management frames the
problem.
However, we don’t want you to get the idea that every linear programming problem falls
entirely and neatly into either one type or the other. In the preceding section and this one, we
are looking at pure resource-allocation problems and pure cost–benefit–trade-off problems.
Although many real problems tend to be either one type or the other, it is fairly common to
have both resource constraints and benefit constraints, even though one may predominate.
(In the next section, you will see an example of how both types of constraints can arise in the
same problem when the management of the Super Grain Corp. introduces additional considerations into the analysis of their advertising-mix problem.) Furthermore, we still need to
consider additional categories of linear programming problems in the remaining sections of
this chapter.
Now, another example of a pure cost–benefit–trade-off problem.
Personnel Scheduling
One of the common applications of cost–benefit–trade-off analysis involves personnel scheduling for a company that provides some kind of service, where the objective is to schedule
the work times of the company’s employees so as to minimize the cost of providing the level
of service specified by management. The following example illustrates how this can be done.
The Problem
Union Airways is adding more flights to and from its hub airport and so needs to hire additional customer service agents. However, it is not clear just how many more should be hired.
Management recognizes the need for cost control while also consistently providing a satisfactory level of service to the company’s customers, so a desirable trade-off between these two
factors is being sought. Therefore, a management science team is studying how to schedule
the agents to provide satisfactory service with the smallest personnel cost.
Based on the new schedule of flights, an analysis has been made of the minimum number
of customer service agents that need to be on duty at different times of the day to provide a
satisfactory level of service. (The queueing models presented in Chapter 11 can be used to
determine the minimum numbers of agents needed to reduce the waiting times incurred by
customers down to reasonable levels.) These numbers are shown in the last column of Table
3.5 for the time periods given in the first column. The other entries in this table reflect one of
the provisions in the company’s current contract with the union that represents the customer
service agents. The provision is that each agent works an eight-hour shift. The authorized
shifts are
Shift 1: 6:00 AM to 2:00 PM.
Shift 2: 8:00 AM to 4:00 PM.
Shift 3: Noon to 8:00 PM.
Shift 4: 4:00 PM to midnight.
Shift 5: 10:00 PM to 6:00 AM.
84 Chapter Three Linear Programming: Formulation and Applications
TABLE 3.5
Data for the Union Airways Personnel Scheduling Problem
Time Periods Covered by Shift
Time Period
1
6:00 AM to 8:00 AM
8:00 AM to 10:00 AM
10:00 AM to noon
Noon to 2:00 PM
2:00 PM to 4:00 PM
4:00 PM to 6:00 PM
6:00 PM to 8:00 PM
8:00 PM to 10:00 PM
10:00 PM to midnight
Midnight to 6:00 AM
✓
✓
✓
✓
Daily cost per agent
$170
2
✓
✓
✓
✓
$160
3
✓
✓
✓
✓
$175
4
✓
✓
✓
✓
$180
5
Minimum Number
of Agents Needed
✓
✓
48
79
65
87
64
73
82
43
52
15
$195
Check marks in the main body of Table 3.5 show the time periods covered by the respective shifts. Because some shifts are less desirable than others, the wages specified in the
contract differ by shift. For each shift, the daily compensation (including benefits) for each
agent is shown in the bottom row. The problem is to determine how many agents should be
assigned to the respective shifts each day to minimize the total personnel cost for agents,
based on this bottom row, while meeting (or surpassing) the service requirements given in
the last column.
Formulation
This problem is, in fact, a pure cost–benefit–trade-off problem. To formulate the problem, we
need to identify the activities and benefits involved.
Activities correspond to shifts.
The level of each activity is the number of agents assigned to that shift.
A unit of each activity is one agent assigned to that shift.
Thus, the general description of a linear programming problem as finding the best mix of
activity levels can be expressed for this specific application as finding the best mix of shift
sizes.
Benefits correspond to time periods.
For each time period, the benefit provided by the activities is the service that agents provide customers during that period.
The level of a benefit is measured by the number of agents on duty during that time period.
Once again, a careful formulation of the problem, including gathering all the relevant data,
leads rather directly to a spreadsheet model. This model is shown in Figure 3.5, and we outline its formulation below.
The Data As indicated in this figure, all the data in Table 3.5 have been entered directly
into the data cells CostPerShift (C5:G5), ShiftWorksTimePeriod (C8:G17), and MinimumNeeded (J8:J17). For the ShiftWorksTimePeriod (C8:G17) data, an entry of 1 indicates that
the corresponding shift includes that time period whereas 0 indicates not. Like any cost–­
benefit–tradeoff problem, these numbers indicate the contribution of each activity to each
benefit. Each agent working a shift contributes either 0 or 1 toward the minimum number of
agents needed in a time period.
The Decisions Since the activities in this case correspond to the five shifts, the decisions to
be made are
S1 = Number of agents to assign to Shift 1 (starts at 6 AM)
S2 = Number of agents to assign to Shift 2 (starts at 8 AM)
3.3 Cost–Benefit–Trade-Off Problems 85
FIGURE 3.5
The spreadsheet model for the Union Airways problem, including the formulas for the objective cell TotalCost (J21) and the other
output cells TotalWorking (H8:H17), as well as the specifications needed to set up Solver. The changing cells NumberWorking
(C21:G21) show the optimal solution obtained by Solver.
A
1
B
D
C
E
F
G
H
I
J
Union Airways Personnel Scheduling Problem
2
3
6AM–2PM
8AM–4PM
Noon–8PM
4PM–Midnight
10PM–6AM
4
Shift
Shift
Shift
Shift
Shift
$170
$160
$175
$180
$195
5
Cost per Shift
6
Total
Minimum
Working
7
Time Period
8
6AM–8AM
1
0
0
0
0
9
8AM–10AM
1
1
0
0
0
Shift Works Time Period? (1=yes, 0=no)
Needed
48
≥
48
79
≥
79
10
10AM–12PM
1
1
0
0
0
79
≥
65
11
12PM–2PM
1
1
1
0
0
118
≥
87
12
2PM–4PM
0
1
1
0
0
70
≥
64
13
4PM–6PM
0
0
1
1
0
82
≥
73
14
6PM–8PM
0
0
1
1
0
82
≥
82
15
8PM–10PM
0
0
0
1
0
43
≥
43
16
10PM–12AM
0
0
0
1
1
58
≥
52
15
≥
15
17
12AM–6AM
0
0
0
0
1
6AM–2PM
8AM–4PM
Noon–8PM
4PM–Midnight
10PM–6AM
Shift
Shift
Shift
Shift
Shift
Total Cost
48
31
39
43
15
$30,610
18
19
20
21
Number Working
Solver Parameters
Set Objective Cell: TotalCost
To: Min
By Changing Variable Cells:
NumberWorking
Subject to the Constraints:
NumberWorking = integer
TotalWorking >= MinimumNeeded
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
Range Name
Cells
CostPerShift
MinimumNeeded
NumberWorking
ShiftWorksTimePeriod
TotalCost
TotalWorking
C5:G5
J8:J17
C21:G21
C8:G17
J21
H8:H17
H
6
Total
7
Working
8
9
10
=SUMPRODUCT(C8:G8,NumberWorking)
=SUMPRODUCT(C9:G9,NumberWorking)
=SUMPRODUCT(C10:G10,NumberWorking)
11
=SUMPRODUCT(C11:G11,NumberWorking)
12
=SUMPRODUCT(C12:G12,NumberWorking)
13
14
15
16
17
=SUMPRODUCT(C13:G13,NumberWorking)
=SUMPRODUCT(C14:G14,NumberWorking)
=SUMPRODUCT(C15:G15,NumberWorking)
=SUMPRODUCT(C16:G16,NumberWorking)
=SUMPRODUCT(C17:G17,NumberWorking)
J
20
Total Cost
21
=SUMPRODUCT(CostPerShift,NumberWorking)
S3 = Number of agents to assign to Shift 3 (starts at noon)
S4 = Number of agents to assign to Shift 4 (starts at 4 PM)
S5 = Number of agents to assign to Shift 5 (starts at 10 PM)
The changing cells to hold these numbers have been placed in the activity columns in row 21, so
​​​S​  1​​ → cell C21​
​S​  2​​ → cell D21​
. . .​
​S​  5​​ → cell G21​​
86 Chapter Three Linear Programming: Formulation and Applications
where these cells are collectively referred to by the range name NumberWorking (C21:G21).
The Constraints These changing cells need to be nonnegative. In addition, we need 10
benefit constraints, where each one specifies that the total number of agents serving in the
corresponding time period listed in column B must be no less than the minimum acceptable
number given in column J. Thus, these constraints are
Total number of agents serving 6 − 8 am
≥ 48
(min. acceptable)
Total number of agents serving 8 − 10 am
≥ 79
(min. acceptable)
​
.
​
​
​​​
​  ​
​
​  ​  ​  ​
​
​​​
​
. ​
​
​
. ​
​
Total number of agents serving midnight − 6 am ≥ 15
(min. acceptable)
Since columns C to G indicate which of the shifts serve each of the time periods, these totals are
Total number of agents serving 6 − 8 am
= ​S​  1​​
Total number of agents serving 8 − 10 am
= ​S​  1​​ + ​S​  2​​
​
.​  ​  ​​  ​​
​​​
​  ​
​
​
​
​
. ​
​
. ​
Total number of agents serving midnight − 6 am = ​S​  5​​
These totals are calculated in the output cells TotalWorking (H8:H17) using the SUMPRODUCT functions shown below the spreadsheet in Figure 3.5.
One other type of constraint is that the number of agents assigned to each shift must have
an integer value. These constraints for the five shifts should be added in the same way as
described for the TBA Airlines problem in Section 3.2. In particular, with Excel’s Solver they
are added in the Add Constraint dialog box by entering NumberWorking on the left-hand side
and then choosing int from the pop-up menu between the left-hand side and the right-hand
side. The set of constraints, NumberWorking = integer, then appears in the Solver Parameters,
as shown in Figure 3.5. In Analytic Solver, select the range of cells to be constrained integer,
and then under the Constraint menu on the Analytic Solver ribbon, choose Integer under the
Variable Type/Bound submenu.
The Measure of Performance The objective is to
​​Minimize
​  Cost = Total daily personnel cost for all agents​​
Since CostPerShift (C5:G5) gives the daily cost per agent on each shift and NumberWorking
(C21:G21) gives the number of agents working each shift,
Cost = 170 ​S​  1​​ + 160 ​S​  2​​ + 175 ​S​  3​​ + 180 ​S​  4​​ + 195 ​S​  5​​ (in dollars)
​ ​
​​
​  =​  ​  SUMPRODUCT (CostPerShift, NumberWorking)​
​​​
​
→ cell J21
Summary of the Formulation The above steps provide the complete formulation of the
linear programming model on a spreadsheet, as summarized below (in algebraic form).
​​Minimize​  Cost = 170 ​S​  1​​ + 160 ​S​  2​​ + 175 ​S​  3​​ + 180 ​S​  4​​ + 195 ​S​  5​​​
subject to
Total agents 6 − 8 am
​ ​S​  1​​
≥ 48
Total agents 8 − 10 am
​ ​S​  1​​ + ​S​  2​​
≥ 79
.
​
​
​​
​
​  ​
​  ​  ​  ​
​
​  ​​  ​  ​​ ​​​
.
​ ​
​ ​
.
​ ​
​ ​
Total agents midnight − 6 am ​
​ S​  5​​ ≥ 15
(in dollars)​
3.4 Mixed Problems 87
FIGURE 3.6
A template of a spreadsheet model for pure cost–benefit–trade-off problems.
Unit Cost
Cost per unit of activity
Constraints
Activities
Benefit achieved per unit of activity
Benefit
Achieved
Benefit
Needed
SUMPRODUCT
(benefit per unit,
changing cells)
≥
Total Cost
Level of Activity
Changing cells
SUMPRODUCT(cost per unit, changing cells)
and
​​​S​  1​​ ≥ 0​
​S​  2​​ ≥ 0​
​S​  3​​ ≥ 0​
​S​  4​​ ≥ 0​
​S​  5​​ ≥ 0​​
Solving the Model The lower left-hand corner of Figure 3.5 shows the entries needed in Solver,
along with the selection of the usual two options. After solving, NumberWorking (C21:G21) in
the spreadsheet shows the resulting optimal solution for the number of agents that should be
assigned to each shift. TotalCost (J21) indicates that this plan would cost $30,610 per day.
Summary of the Formulation Procedure for Cost–Benefit–Trade-Off
Problems
The nine steps in formulating any cost–benefit–trade-off problem follow the same pattern as
presented at the end of the preceding section for resource-allocation problems, so we will not
repeat them here. The main differences are that the overall measure of performance now is
the total cost of the activities (or some surrogate of total cost chosen by management) in steps
2 and 3, benefits now replace resources in steps 4 and 5, and ≥ signs now are entered to the
right of the output cells for benefits in step 8. Figure 3.6 shows a template of the format of a
spreadsheet model for cost–benefit–trade-off problems.
All the functional constraints in the resulting model are benefit constraints, that is, constraints with a ≥ sign. This is the identifying feature of a pure cost–benefit–trade-off problem.
Review
Questions
3.4
1. What is the difference in managerial objectives between resource-allocation problems and
cost–benefit–trade-off problems?
2. What is the identifying feature of a cost–benefit–trade-off problem?
3. What is the form of a benefit constraint?
4. What are the three kinds of data that need to be gathered for a cost–benefit–trade-off problem?
5. Compare the types of activities for the two examples of cost–benefit–trade-off problems.
6. Compare the types of benefits for the two examples of cost–benefit–trade-off problems.
MIXED PROBLEMS
Sections 3.2 and 3.3 each described a broad category of linear programming problems—
resource-allocation and cost–benefit–trade-off problems. As summarized in Table 3.6, each
features one of the first two types of functional constraints shown there. In fact, the identifying
feature of a pure resource-allocation problem is that all its functional constraints are resource
constraints. The identifying feature of a pure cost–benefit–trade-off problem is that all its
functional constraints are benefit constraints. (Keep in mind that the functional constraints
include all the constraints of a problem except its nonnegativity constraints.)
88 Chapter Three Linear Programming: Formulation and Applications
The bottom row of Table 3.6 shows the last of the three types of functional constraints,
namely, fixed-requirement constraints, which require that the left-hand side of each such
constraint must exactly equal some fixed amount. Thus, since the left-hand side represents the
amount provided of some quantity, the form of a fixed-requirement constraint is
​Amount provided = Required amount​
The identifying feature of a pure fixed-requirements problem is that it is a linear programming problem where all its functional constraints are fixed-requirement constraints. The next
two sections will describe two particularly prominent types of fixed-requirement problems
called transportation problems and assignment problems.
However, before turning to these types of problems, we first will use a continuation of the
Super Grain case study from Section 3.1 to illustrate how many linear programming problems
fall into another broad category called mixed problems.
Many linear programming problems do not fit completely into any of the previously discussed
categories (pure resource-allocation problems, cost–benefit–trade-off problems, and fixedrequirement problems) because the problem’s functional constraints include more than one of
the types shown in Table 3.6. Such problems are called mixed problems.
Now let us see how a more careful analysis of the Super Grain case study turns this resourceallocation problem into a mixed problem that includes all three types of functional constraints
shown in Table 3.6.
Super Grain Management Discusses Its Advertising-Mix Problem
The description of the Super Grain case study in Section 3.1 ends with Clair Syverson (Super
Grain’s vice president for marketing) sending a memorandum to the company’s president,
David Sloan, requesting a meeting to evaluate her proposed promotional campaign for the
company’s new breakfast cereal.
Soon thereafter, Claire Syverson and David Sloan meet to discuss plans for the campaign.
David Sloan (president): Thanks for your memo, Claire. The plan you outline for the
promotional campaign looks like a reasonable one. However, I am surprised that it does not
make any use of TV commercials. Why is that?
Claire Syverson (vice president for marketing): Well, as I described in my memo, I used
a spreadsheet model to see how to maximize the number of exposures from the campaign
and this turned out to be the plan that does this. I also was surprised that it did not include
TV commercials, but the model indicated that introducing commercials would provide less
exposures on a dollar-for-dollar basis than magazine ads and Sunday supplement ads. Don’t
you think it makes sense to use the plan that maximizes the number of exposures?
David: Not necessarily. Some exposures are a lot less important than others. For example,
we know that middle-aged adults are not big consumers of our cereals, so we don’t care
very much how many of those people see our ads. On the other hand, young children are big
consumers. Having TV commercials on the Saturday morning programs for children is our
TABLE 3.6
Types of Functional Constraints
Type
Form*
Typical Interpretation
Main Usage
Resource constraint
LHS ≤ RHS
Benefit constraint
LHS ≥ RHS
Resource-allocation problems
and mixed problems
Cost–benefit–trade-off problems and mixed problems
Fixed-requirement
constraint
LHS = RHS
For some resource,
Amount used ≤ Amount available
For some benefit,
Level achieved ≥ Minimum
acceptable level
For some quantity,
Amount provided = Required amount
*LHS = Left-hand side (a SUMPRODUCT function).
RHS = Right-hand side (a constant).
Fixed-requirements problems
and mixed problems
3.4 Mixed Problems 89
primary method of reaching young children. You know how important it will be to get young
children to ask their parents for Crunchy Start. That is our best way of generating first-time
sales. Those commercials also get seen by a lot of parents who are watching the programs
with their kids. What we need is a commercial that is appealing to both parents and kids, and
that gets the kids immediately bugging their parents to go buy Crunchy Start. I think that is a
real key to a successful campaign.
Claire: Yes, that makes a lot of sense. In fact, I already have set some goals regarding
the number of young children and the number of parents of young children that need to be
reached by this promotional campaign.
David: Good. Did you include those goals in your spreadsheet model?
Claire: No, I didn’t.
David: Well, I suggest that you incorporate them directly into your model. I suspect that
maximizing exposures while also meeting your goals will give us a high impact plan that
includes some TV commercials.
Claire: Good idea. I’ll try it.
David: Are there any other factors that the plan in your memo doesn’t take into account as
well as you would like?
Claire: Well, yes, one. The plan doesn’t take into account my budget for cents-off coupons
in magazines and newspapers.
David: You should be able to add that to your model as well. Why don’t you go back and
see what happens when you incorporate these additional considerations?
Claire: OK, will do. You seem to have had a lot of experience with spreadsheet
modeling.
David: Yes. It is a great tool as long as you maintain some healthy skepticism about what
comes out of the model. No model can fully take into account everything that we must consider when dealing with managerial problems. This is especially true the first time or two
you run the model. You need to keep asking, what are the missing quantitative considerations that I still should add to the model? Then, after you have made the model as complete
as possible and obtained a solution, you still need to use your best managerial judgment to
weigh intangible considerations that cannot be incorporated into the model.
Incorporating Additional Managerial Considerations
into the Super Grain Model
Therefore, David and Claire conclude that the spreadsheet model needs to be expanded to
incorporate some additional considerations. In particular, since the promotional campaign
is for a breakfast cereal that should have special appeal to young children, they feel that two
audiences should be targeted—young children and parents of young children. (This is why
one of the three advertising media recommended by Giacomi & Jackowitz is commercials on
children’s television programs Saturday morning.) Consequently, Claire now has set two new
goals for the campaign.
Goal 1: The advertising should be seen by at least five million young children.
Goal 2: The advertising should be seen by at least five million parents of young children.
In effect, these two goals are minimum acceptable levels for two special benefits to be
achieved by the advertising activities.
Benefit 1: Promoting the new breakfast cereal to young children.
Benefit 2: Promoting the new breakfast cereal to parents of young children.
Because of the way the goals have been articulated, the level of each of these benefits is measured by the number of people in the specified category that are reached by the advertising.
To enable constructing the corresponding benefit constraints (as described in Section 3.3),
Claire asks Giacomi & Jackowitz to estimate how much each advertisement in each of the
media will contribute to each benefit, as measured by the number of people reached in the
specified category. These estimates are given in Table 3.7.
90 Chapter Three Linear Programming: Formulation and Applications
TABLE 3.7
Benefit Data for the Revised Super Grain Corp. Advertising-Mix Problem
Number Reached in Target Category (in millions)
Each TV
Commercial
Each
Magazine Ad
Young children
1.2
0.1
0
5
Parents of young
children
0.5
0.2
0.2
5
Target Category
Benefit constraints are useful for incorporating managerial goals into the model.
Each
Sunday Ad
Minimum
Acceptable Level
It is interesting to observe that management wants special consideration given to these two
kinds of benefits even though the original spreadsheet model (Figure 3.1) already takes them
into account to some extent. As described in Section 3.1, the expected number of exposures
is the overall measure of performance to be maximized. This measure counts up all the times
that an advertisement is seen by any individual, including all those individuals in the target
audiences. However, maximizing this general measure of performance does not ensure that
the two specific goals prescribed by management (Claire Syverson) will be achieved. Claire
feels that achieving these goals is essential to a successful promotional campaign. Therefore,
she complements the general objective with specific benefit constraints that do ensure that the
goals will be achieved. Having benefit constraints added to incorporate managerial goals into
the model is a prerogative of management.
Claire has one more consideration she wants to incorporate into the model. She is a strong
believer in the promotional value of cents-off coupons (coupons that shoppers can clip from
printed advertisements to obtain a refund of a designated amount when purchasing the advertised item). Consequently, she always earmarks a major portion of her annual marketing budget for the redemption of these coupons. She still has $1,490,000 left from this year’s allotment
for coupon redemptions. Because of the importance of Crunchy Start to the company, she has
decided to use this entire remaining allotment in the campaign promoting this cereal.
This fixed amount for coupon redemptions is a fixed requirement that needs to be expressed
as a fixed-requirement constraint. As described at the beginning of this section, the form of a
fixed-requirement constraint is that, for some type of quantity,
​Amount provided = Required amount​
In this case, the quantity involved is the amount of money provided for the redemption of
cents-off coupons. To specify this constraint in the spreadsheet, we need to estimate how
much each advertisement in each of the media will contribute toward fulfilling the required
amount for the quantity. Both medium 2 (advertisements in food and family-oriented magazines) and medium 3 (advertisements in Sunday supplements of major newspapers) will feature cents-off coupons. The estimates of the amount of coupon redemption per advertisement
in each of these media is given in Table 3.8.
Formulation of the Revised Spreadsheet Model
Figure 3.7 shows one way of formatting the spreadsheet to expand the original spreadsheet
model in Figure 3.1 to incorporate the additional managerial considerations. We then outline
the four components of the revised model next.
TABLE 3.8
Data for the Fixed-Requirement Constraint for the Revised Super Grain Corp. Advertising-Mix Problem
Contribution toward Required Amount
Requirement
Coupon redemption
Each
TV Spot
Each
Magazine Ad
Each
Sunday Ad
Required
Amount
0
$40,000
$120,000
$1,490,000
3.4 Mixed Problems 91
FIGURE 3.7
The spreadsheet model for the revised Super Grain problem, including the formulas for the objective cell TotalExposures (H19) and
the other output cells in column F, as well as the specifications needed to set up Solver. The changing cells NumberOfAds (C19:E19)
show the optimal solution obtained by Solver.
A
1
B
C
D
E
F
TV Spots
Magazine Ads
SS Ads
1,300
600
500
G
H
Super Grain Corp. Advertising-Mix Problem
2
3
4
Exposures per Ad
5
(thousands)
Cost per Ad ($thousands)
6
Budget Spent
Budget Available
Ad Budget
300
150
100
3,775
≤
4,000
Planning Budget
90
30
40
1,000
≤
1,000
11
Young Children
1.2
0.1
0
5
≥
12
Parents of Young Children
0.5
0.2
0.2
5.85
≥
TV Spots
Magazine Ads
SS Ads
Total Redeemed
0
40
120
1,490
7
8
9
10
Number Reached per Ad (millions)
Total Reached
Minimum Acceptable
5
TotalExposures
5
($thousands)
13
14
15
Coupon Redemption
16
per Ad ($thousands)
Required Amount
=
1,490
17
Total Exposures
18
19
Number of Ads
Magazine Ads
SS Ads
(thousands)
3
14
7.75
16,175
≤
20
21
TV Spots
Maximum TV Spots
Solver Parameters
Set Objective Cell: TotalExposures
To: Max
By Changing Variable Cells:
NumberOfAds
Subject to the Constraints:
BudgetSpent <= Budget Available
TVSpots <= MaxTVSpots
TotalReached >= MinimumAcceptable
TotalRedeemed = RequiredAmount
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
5
Range Name
Cells
BudgetAvailable
BudgetSpent
CostPerAd
CouponRedemptionPerAd
ExposuresPerAd
MaxTVSpots
MinimumAcceptable
NumberOfAds
NumberReachedPerAd
RequiredAmount
TotalExposures
TotalReached
TotalRedeemed
TVSpots
H7:H8
F7:F8
C7:E8
C15:E15
C4:E4
C21
H11:H12
C19:E19
C11:E12
H15
H19
F11:F12
F15
C19
F
6
Budget Spent
7
=SUMPRODUCT(C7:E7,NumberOfAds)
8
9
=SUMPRODUCT(C8:E8,NumberOfAds)
10
Total Reached
11
=SUMPRODUCT(C11:E11,NumberOfAds)
12
13
=SUMPRODUCT(C12:E12,NumberOfAds)
14
15
Total Redeemed
=SUMPRODUCT(CouponRedemptionPerAd,
NumberOfAds)
H
17
18
19
Total Exposures
(thousands)
=SUMPRODUCT(ExposuresPerAd,NumberOfAds)
The Data
Additional data cells in NumberReachedPerAd (C11:E12), MinimumAcceptable (H11:H12),
CouponRedemptionPerAd (C15:E15), and RequiredAmount (H15) give the data in Tables 3.7
and 3.8.
92 Chapter Three Linear Programming: Formulation and Applications
The Decisions
Recall that, as before, the decisions to be made are
TV = Number of commercials on television
M = Number of advertisements in magazines
SS = Number of advertisements in Sunday supplements
The changing cells to hold these numbers continue to be in NumberOfAds (C19:E19).
The Constraints
In addition to the original constraints, we now have two benefit constraints and one fixedrequirement constraint. As specified in rows 11 and 12, columns F to H, the benefit constraints are
Total number of young children reached ≥ 5
(goal 1 in millions)
​​
​
​
​
​​​
Total number of parents reached
≥5
(goal 2 in millions)
Using the data in columns C to E of these rows,
Total number of young children reached = 1.2TV + 0.1M + 0SS
​
= SUMPRODUCT (C11:E11, NumberOfAds)
​​​
→
​
​
​
​  ​  cell F11​
​ ​
​​​
Total number of parents reached
= 0.5TV + 0.2M + 0.2SS
​
= SUMPRODUCT (C12:E12, NumberOfAds)
​
→ cell F12
These output cells are given the range name TotalReached (F11:F12).
The fixed-requirement constraint indicated in row 15 is that
​​Total coupon redemption = 1,490​
​​(​​allotment in $1,000s​)​​​​​
CouponRedemptionPerAd (C15:E15) gives the number of coupons redeemed per ad, so
Total coupon redemption = 0TV + 40M + 120SS
​​
​
​  =​  ​  SUMPRODUCT (CouponRedemptionPerAd,
​
NumberOfAds)​​​
​
→ cell F15
These same constraints are specified in Solver, along with the original constraints, in
Figure 3.7.
The Measure of Performance
The measure of performance continues to be
Exposure = 1,300TV + 600M + 500SS
​​
​
​  =​  ​  SUMPRODUCT (ExposuresPerAd,
​
​
NumberOfAds)​​​
​
→ cell H19
so the objective cell is again TotalExposures (H19).
Summary of the Formulation
The above steps have resulted in formulating the following linear programming model (in
algebraic form) on a spreadsheet.
​​Maximize ​  Exposure = 1,300TV + 600M + 500SS​​
subject to the following constraints:
1. Resource constraints:
300TV + 150M + 100SS ≤ 4,000
​​(​​ad budget in $1,000s​)​​​
​​
​  ≤​ ​  1,000​ ​
​  ​​)​​​​
90TV ​  + 30M +
40SS​
(​​ ​​planning budget in $1,000s​
TV
≤    5
(​​ ​​television spots available​)​​​
An Application Vignette
The Chevron Corporation is one of the world’s leading integrated energy companies. It explores extensively for crude
oil and natural gas throughout the world. Thanks to its vast
reserves, it produces nearly 2 million barrels of crude oil per
day and similar amounts of natural gas. It then uses its refineries to refine and market nearly 3 million barrels per day of transportation fuels, chemicals, and lubricants.
Dating nearly back to the invention of linear programming
in 1947, Chevron quickly became one of the heaviest users of
this exciting new technique. The earliest applications involved
the following blending problem. Any single grade of gasoline
needs to be blended from about three to ten components (different forms of processed crude oil), where no single component meets the quality specifications of the grade of gasoline
but various combinations of the components can accomplish
this. A typical refinery might have 20 different components
to be blended into four or more grades of gasoline differing
in octane and other properties by marketing area. Linear programming of the mixed type can achieve huge savings by solving for how to minimize the total cost of accomplishing all this
blending.
As time went on, the exponential growth in computing
power enabled Chevron to greatly expand its use of linear
programming. One application involves optimizing the combination of refined products (gasoline, jet, diesel fuels) to produce to maximize the total profit. Another application involves
periodically determining the optimal way to run the refining
processing units when changes occur in crude oil prices, raw
material availability, product prices, product specifications, and
equipment capabilities. Still another application of linear programming (along with also using decision analysis, the subject
of Chapter 9) involves optimizing the use of capital for new projects to improve its refining system on an ongoing basis.
The combination of all these applications of linear programming to minimize the total cost or maximize the total profit in
various ways has had a dramatic impact on Chevron’s bottom line. The estimated cumulative value to Chevron now
approaches $1 billion annually. In recognition of this and
related work, the Institute for Operations Research and the
Management Sciences (INFORMS) awarded Chevron the prestigious 2015 INFORMS Prize for its long and innovative history in
applying advanced analytics and management science across
the company.
Source: T. Kutz, M. Davis, R. Creek, N. Kenaston, C. Stenstrom, and
M. Connor, Interfaces 44, no. 1 (January-February 2014), pp. 39–54.
(A link to this article is available at www.mhhe.com/Hillier6e.)
2. Benefit constraints:
1.2TV + 0.1M
≥5
​​(​​millions of young children​)​​​
​​
​
​
​
​​​
0.5TV + 0.2M + 0.2SS ≥ 5
​​(​​millions of parents​)​​​
3. Fixed-requirement constraint:
​​40M + 120SS = 1,490​
​​(​​coupon budget in $1,000s​)​​​​​
4. Nonnegativity constraints:
​​ V ≥ 0​
T
M ≥ 0​
SS ≥ 0​​
Solving the Model
The lower left-hand corner of Figure 3.7 shows the entries needed in Solver, along with the
selection of the usual two options. Solver then finds the optimal solution given in row 19. This
optimal solution provides the following plan for the promotional campaign:
Run 3 television commercials.
Run 14 advertisements in magazines.
Run 7.75 advertisements in Sunday supplements (so the eighth advertisement would
appear in only 75 percent of the newspapers).
A model may need to be
modified a number of times
before it adequately incorporates all the important
considerations.
Although the expected number of exposures with this plan is only 16,175,000, versus the
17,000,000 with the first plan shown in Figure 3.1, both Claire Syverson and David Sloan feel
that the new plan does a much better job of meeting all of management’s goals for this campaign. They decide to adopt the new plan.
This case study illustrates a common theme in real applications of linear programming—
the continuing evolution of the linear programming model. It is common to make later adjustments in the initial version of the model, perhaps even many times, as experience is gained
in using the model. Frequently, these adjustments are made to more adequately reflect some
important managerial considerations. This may result in a mixed problem because the new
functional constraints needed to incorporate the managerial considerations may be of a different type from those in the original model.
93
94 Chapter Three Linear Programming: Formulation and Applications
Other Examples
This case study provides a relatively simple example of a small mixed problem. Most of
the mixed problems that arise in practice are much larger, sometimes involving hundreds
or thousands of activities and hundreds or thousands of constraints. At first glance, these
larger problems may seem considerably more complicated than the case study. However,
the important thing to remember is that any linear programming problem can have only
three types of functional constraints—resource constraints, benefit constraints, and fixedrequirement ­constraints—where each type is formulated just as illustrated above for the
case study.
There are numerous kinds of managerial problems to which linear programming can be
applied. We don’t have nearly enough space available to give examples of all the most important kinds of applications. However, if you would like to explore this further, we suggest that
you go through the five solved problems that are summarized in front of the Problems section
for this chapter. Reading the seven cases that follow the Problems section, as well as the application vignettes in both this chapter and the preceding chapter, also will further illustrate the
unusually wide applicability of linear programming.
Meanwhile, we soon will turn to two more categories of linear programming problems in
the next two sections.
Summary of the Formulation Procedure for Mixed
Linear Programming Problems
The procedure for formulating mixed problems is similar to the one outlined at the end of
Section 3.2 for resource-allocation problems. However, pure resource-allocation prob­
lems only have resource constraints whereas mixed problems can include all three types of
functional constraints (resource constraints, benefit constraints, and fixed-requirement constraints). Therefore, the following summary for formulating mixed problems includes separate
steps for dealing with these different types of constraints. Also see Figure 3.8 for a template
of the format for a spreadsheet model of mixed problems. (This format works well for most
mixed problems, including those encountered in this chapter, but more flexibility is occasionally needed, as will be illustrated in the next chapter.)
FIGURE 3.8
A template of a spreadsheet model for mixed problems.
Activities
Unit Profit or Cost
Profit/cost per unit of activity
Constraints
Resource used per unit of activity
SUMPRODUCT
(resource used per unit,
changing cells)
Resources
Available
≤
Benefit
Achieved
Benefit achieved per unit of activity
Level of Activity
Resources
Used
Changing cells
SUMPRODUCT
(benefit per unit,
changing cells)
Benefit
Needed
≥
=
Total Profit or Cost
SUMPRODUCT(profit/cost per unit, changing cells)
3.5 Transportation Problems 95
1. Since any linear programming problem involves finding the best mix of levels of various
activities, identify these activities for the problem at hand. The decisions to be made are
the levels of these activities.
2. From the viewpoint of management, identify an appropriate overall measure of performance for solutions of the problem.
3. For each activity, estimate the contribution per unit of the activity to this overall measure
of performance.
4. Identify any resources that must be allocated to the activities (as described in Section 3.2). For
each one, identify the amount available and then the amount used per unit of each activity.
5. Identify any benefits to be obtained from the activities (as described in Section 3.3). For
each one, identify the minimum acceptable level prescribed by management and then the
benefit contribution per unit of each activity.
6. Identify any fixed requirements that, for some type of quantity, the amount provided must
equal a required amount (as described in Section 3.4). For each fixed requirement, identify the required amount and then the contribution toward this required amount per unit
of each activity.
7. Enter the data gathered in steps 3–6 into data cells in a spreadsheet.
8. Designate changing cells for displaying the decisions on activity levels.
9. Use output cells to specify the constraints on resources, benefits, and fixed requirements.
10. Designate an objective cell for displaying the overall measure of performance.
Review
Questions
3.5
1. What types of functional constraints can appear in a mixed linear programming problem?
2. What managerial goals needed to be incorporated into the expanded linear programming
model for the Super Grain Corp. problem?
3. Which categories of functional constraints are included in the new linear programming model?
4. Why did management adopt the new plan even though it provides a smaller expected number
of exposures than the original plan recommended by the original linear programming model?
TRANSPORTATION PROBLEMS
One of the most common applications of linear programming involves optimizing a shipping
plan for transporting goods. In a typical application, a company has several plants producing
a certain product that needs to be shipped to the company’s customers (or perhaps to distribution centers). How much should each plant ship to each customer in order to minimize the
total cost? Linear programming can provide the answer. This type of linear programming
problem is called a transportation problem.
This kind of application normally needs two kinds of functional constraints. One kind
specifies that the amount of the product produced at each plant must equal the total amount
shipped to customers. The other kind specifies that the total amount received from the plants
by each customer must equal the amount ordered. These are fixed-requirement constraints,
which makes the problem a fixed-requirements problem. However, there also are variations of
this problem where resource constraints or benefit constraints are needed.
Transportation problems and assignment problems (described in the next section) are
such important types of linear programming problems that the entire Chapter 15 (available at
www.mhhe.com/Hillier6e) is devoted to further describing these two related types of problems and providing examples of a wide variety of applications.
We provide below an example of a typical transportation problem.
The Big M Company Transportation Problem
The Big M Company produces a variety of heavy duty machines at two factories. One of its
products is a large turret lathe. Orders have been received from three customers to purchase
some of these turret lathes next month. These lathes will be shipped individually, and Table 3.9
shows what the cost will be for shipping each lathe from each factory to each customer. This
96 Chapter Three Linear Programming: Formulation and Applications
TABLE 3.9
Some Data for the Big M
Company DistributionNetwork Problem
Shipping Cost for Each Lathe
To
Customer 1
Customer 2
Customer 3
Output
From
Factory 1
Factory 2
$700
800
$900
900
$800
700
12 lathes
15 lathes
Order size
10 lathes
8 lathes
9 lathes
table also shows how many lathes have been ordered by each customer and how many will be
produced by each factory. The company’s distribution manager now wants to determine how
many machines to ship from each factory to each customer to minimize the total shipping cost.
Figure 3.9 depicts the distribution network for this problem. This network ignores the geographical layout of the factories and customers and instead lines up the two factories in one
column on the left and the three customers in one column on the right. Each arrow shows one
of the shipping lanes through this distribution network.
Formulation of the Problem in Linear Programming Terms
We need to identify the activities and requirements of this transportation problem to formulate it as a linear programming problem. In this case, two kinds of activities have been
­mentioned—the production of the turret lathes at the two factories and the shipping of these
lathes along the various shipping lanes. However, we know the specific amounts to be produced at each factory, so no decisions need to be made about the production activities. The
decisions to be made concern the levels of the shipping activities—how many lathes to ship
through each shipping lane. Therefore, we need to focus on the shipping activities for the linear programming formulation.
The activities correspond to shipping lanes, depicted by arrows in Figure 3.9.
The level of each activity is the number of lathes shipped through the corresponding shipping lane.
Just as any linear programming problem can be described as finding the best mix of activity levels, this one involves finding the best mix of shipping amounts for the various shipping
lanes. The decisions to be made are
SF1-C1 = Number of lathes shipped from Factory 1 to Customer 1
SF1-C2 = Number of lathes shipped from Factory 1 to Customer 2
FIGURE 3.9
The distribution network
for the Big M Company
problem.
C1
10 lathes
needed
C2
8 lathes
needed
C3
9 lathes
needed
the
0/la
$70
F1
$900/la
the
$8
00
/la
the
$8
00
/la
th
e
12 lathes
produced
15 lathes
produced
F2
e
/lath
$900
$70
0
/lat
he
3.5 Transportation Problems 97
SF1-C3 = Number of lathes shipped from Factory 1 to Customer 3
SF2-C1 = Number of lathes shipped from Factory 2 to Customer 1
SF2-C2 = Number of lathes shipped from Factory 2 to Customer 2
SF2-C3 = Number of lathes shipped from Factory 2 to Customer 3
so six changing cells will be needed in the spreadsheet.
The objective is to
​​Minimize
​  Cost = Total cost for shipping the lathes​​
Using the shipping costs given in Table 3.9,
​Cost = 700 ​S​  F1-C1​​ + 900 ​S​  F1-C2​​ + 800 ​S​  F1-C3​​ + 800 ​S​  F2-C1​​ + 900 ​S​  F2-C2​​ + 700 ​S​  F2-C3​​​
is the quantity in dollars to be entered into the objective cell. (We will use a SUMPRODUCT
function to do this a little later.)
The spreadsheet model also will need five constraints involving fixed requirements. Both
Table 3.9 and Figure 3.9 show these requirements.
Requirement 1: Factory 1 must ship 12 lathes.
Requirement 2: Factory 2 must ship 15 lathes.
Requirement 3: Customer 1 must receive 10 lathes.
Requirement 4: Customer 2 must receive 8 lathes.
Requirement 5: Customer 3 must receive 9 lathes.
Thus, there is a specific requirement associated with each of the five locations in the distribution network shown in Figure 3.9.
All five of these requirements can be expressed in constraint form as
​Amount provided = Required amount​
For example, Requirement 1 can be expressed algebraically as
​​S​  F1-C1​​ + ​S​  F1-C2​​ + S​ ​  F1-C3​​ = 12​
where the left-hand side gives the total number of lathes shipped from Factory 1, and 12 is
the required amount to be shipped from Factory 1. Therefore, this constraint restricts SF1-C1,
SF1-C2, and SF1-C3 to values that sum to the required amount of 12. In contrast to the ≤ form
for resource constraints and the ≥ form for benefit constraints, the constraints express fixed
requirements that must hold with equality, so this transportation problem falls into the category
of fixed-requirements problems introduced in the preceding section. However, Chapter 15
(available at www.mhhe.com/Hillier6e) gives several examples that illustrate how variants of
transportation problems can have resource constraints or benefit constraints as well. For example, if 12 lathes represent the manufacturing capacity of Factory 1 (the maximum number that
can be shipped) rather than a requirement for how many must be shipped, the constraint just
given for Requirement 1 would become a ≤ resource constraint instead. Such variations can
be incorporated readily into the spreadsheet model.
Careful problem formulation
needs to precede model
formulation.
Here is an example where
SUM functions are used
for output cells instead of
SUMPRODUCT functions.
Formulation of the Spreadsheet Model
In preparation for formulating the model, the problem has been formulated above by identifying
the decisions to be made, the constraints on these decisions, and the overall measure of performance, as well as gathering all the important data displayed in Table 3.9. All this information
leads to the spreadsheet model shown in Figure 3.10. The data cells include ShippingCost
(C5:E6), Output (H11:H12), and OrderSize (C15:E15), incorporating all the data from Table
3.9. The changing cells are UnitsShipped (C11:E12), which give the decisions on the amounts
to be shipped through the respective shipping lanes. The output cells are TotalShippedOut
(F11:F12) and TotalToCustomer (C13:E13), where the SUM functions entered into these
cells are shown below the spreadsheet in Figure 3.10. The constraints are that TotalShippedOut is required to equal Output and TotalToCustomer is required to equal OrderSize. These
constraints have been specified on the spreadsheet and entered into Solver. The objective
98 Chapter Three Linear Programming: Formulation and Applications
cell is TotalCost (H15), where its SUMPRODUCT function gives the total shipping cost.
The lower left-hand corner of Figure 3.10 shows the entries needed in Solver, along with the
selection of the usual two options.
The layout of the spreadsheet is different than for all the prior linear programming examples
in the book. Rather than a separate column for each activity and a separate row for each constraint, the cost data and changing cells are laid out in a table format. This format provides a
more natural and compact way of displaying the constraints and results.
UnitsShipped (C11:E12) in the spreadsheet in Figure 3.10 shows the result of applying
Solver to obtain an optimal solution for the number of lathes to ship through each shipping
lane. TotalCost (H15) indicates that the total shipping cost for this shipping plan is $20,500.
Since any transportation problem is a special type of linear programming problem, it makes
the standard assumption that fractional solutions are allowed. However, we actually don’t
want this assumption for this particular application since only integer numbers of lathes can
be shipped from a factory to a customer. Fortunately, even while making the standard assumption, the numbers in the optimal solution shown in UnitsShipped (C11:E12) only have integer
values. This is no coincidence. Because of the form of its model, almost any transportation
FIGURE 3.10
The spreadsheet model for the Big M Company problem, including the formulas for the objective cell TotalCost (H15) and the other
output cells TotalShippedOut (F11:F12) and TotalToCustomer (C13:E13), as well as the specifications needed to set up Solver.
The changing cells UnitsShipped (C11:E12) show the optimal solution obtained by Solver.
A
1
B
C
D
E
F
G
H
Big M Company Distribution Problem
2
3
Shipping Cost
4
(per Lathe)
Customer 1
Customer 2
Customer 3
5
Factory 1
$700
$900
$800
6
Factory 2
$800
$900
$700
7
8
Total
9
Shipped
Output
10
Units Shipped
Customer 1
Customer 2
Customer 3
Out
11
Factory 1
12
Factory 2
10
0
2
6
0
9
12
15
13
Total to Customer
10
8
9
=
=
=
Total Cost
10
8
9
$20,500
14
15
Order Size
Solver Parameters
Set Objective Cell: TotalCost
To: Min
By Changing Variable Cells:
UnitsShipped
Subject to the Constraints:
TotalShippedOut = Output
TotalToCustomer = OrderSize
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
13
Range Name
Cells
OrderSize
Output
ShippingCost
TotalCost
TotalShippedOut
TotalToCustomer
UnitsShipped
C15:E15
H11:H12
C5:E6
H15
F11:F12
C13:E13
C11:E12
12
=
=
15
F
8
Total
9
Shipped
10
11
=SUM(C11:E11)
12
=SUM(C12:E12)
Out
B
C
D
E
Total to Customer
=SUM(C11:C12)
=SUM(D11:D12)
=SUM(E11:E12)
H
14
15
Total Cost
=SUMPRODUCT(ShippingCost,UnitsShipped)
3.6 Assignment Problems 99
problem (including this one) is guaranteed in advance to have an optimal solution that has
only integer values despite the fact that fractional solutions also are allowed. In particular, as
long as the data for the problem includes only integer values for all the supplies and demands
(which are the outputs and order sizes in the Big M Company problem), any transportation
problem with feasible solutions is guaranteed to have an optimal solution with integer values
for all its decision variables. Therefore, it is not necessary to add constraints to the model that
require these variables to have only integer values.
To summarize, here is the algebraic form of the linear programming model that has been
formulated in the spreadsheet:
Minimize Cost = 700 SF1-C1 + 900 SF1-C2 + 800 SF1-C3 + 800 SF2-C1
+ 900 SF2-C2 + 700 SF2-C3
subject to the following constraints:
1. Fixed-requirement constraints:
SF1-C1 + SF1-C2 + SF1-C3
= 12 (Factory 1)
SF2-C1 + SF2-C2 + SF2-C3 = 15 (Factory 2)
SF1-C1
+ SF2-C1
SF1-C2
= 10 (Customer 1)
+ SF2-C2
SF1-C3
= 8 (Customer 2)
+ SF2-C3 = 9 (Customer 3)
2. Nonnegativity constraints:
SF1-C1 ≥ 0 SF1-C2 ≥ 0 SF1-C3 ≥ 0 SF2-C1 ≥ 0 SF2-C2 ≥ 0 SF2-C3 ≥ 0
Review
Questions
3.6
1. Why are transportation problems given this name?
2. What is an identifying feature of transportation problems?
3. How does the form of a fixed-requirement constraint differ from that of a resource constraint?
A benefit constraint?
4. What are the quantities with fixed requirements in the Big M Company problem?
ASSIGNMENT PROBLEMS
We now turn to another special type of linear programming problem called assignment
problems. As the name suggests, this kind of problem involves making assignments. Frequently, these are assignments of people to jobs. Thus, many applications of the assignment
problem involve aiding managers in matching up their personnel with tasks to be performed.
Other applications might instead involve assigning machines, vehicles, or plants to tasks.
Here is a typical example.
An Example: The Sellmore Company Problem
The marketing manager of the Sellmore Company will be holding the company’s annual
sales conference soon for sales regional managers and personnel. To assist in the administration of the conference, he is hiring four temporary employees (Ann, Ian, Joan, and Sean),
where each will handle one of the following four tasks:
1.
2.
3.
4.
Word processing of written presentations.
Computer graphics for both oral and written presentations.
Preparation of conference packets, including copying and organizing written materials.
Handling of advance and on-site registrations for the conference.
He now needs to decide which person to assign to each task.
100 Chapter Three Linear Programming: Formulation and Applications
TABLE 3.10
Data for the Sellmore
Co. Problem
Required Time per Task (Hours)
Temporary
Employee
Ann
Ian
Joan
Sean
Decisions need to be made
regarding which person to
assign to each task.
Using Cost (D15:G18),
the objective is to minimize the total cost of the
assignments.
A value of 1 in a changing cell indicates that the
corresponding assignment
is being made, whereas 0
means that the assignment
is not being made.
Excel Tip: When solving an
assignment problem, rounding errors occasionally will
cause Excel to return a noninteger value very close to
0 (e.g., 1.23 E-10, meaning
0.000000000123) or very
close to 1 (e.g., 0.9999912).
To make the spreadsheet
cleaner, you may replace
these “ugly” representations
by their proper value of 0 or
1 in the changing cells.
Word
Processing
35
47
39
32
Graphics
41
45
56
51
Packets
27
32
36
25
Registrations
40
51
43
46
Hourly
Wage
$14
12
13
15
Although each temporary employee has at least the minimal background necessary to perform
any of the four tasks, they differ considerably in how efficiently they can handle the different types
of work. Table 3.10 shows how many hours each would need for each task. The rightmost
column gives the hourly wage based on the background of each employee.
Formulation of a Spreadsheet Model
Figure 3.11 shows a spreadsheet model for this problem. Table 3.10 is entered at the top. Combining these required times and wages gives the cost (cells D15:G18) for each possible assignment
of a temporary employee to a task, using equations shown at the bottom of Figure 3.11. This
cost table is just the way that any assignment problem is displayed. The objective is to determine
which assignments should be made to minimize the sum of the associated costs.
The values of 1 in Supply (J24:J27) indicate that each person (assignee) listed in column
C must perform exactly one task. The values of 1 in Demand (D30:G30) indicate that each
task must be performed by exactly one person. These requirements then are specified in the
constraints given in Solver.
Each of the changing cells Assignment (D24:G27) is given a value of 1 when the corresponding assignment is being made, and a value of 0 otherwise. Therefore, the Excel equation
for the objective cell, TotalCost = SUMPRODUCT(Cost, Assignment), gives the total cost for
the assignments being made. The Solver Parameters box specifies that the goal is to minimize
this objective cell.
The changing cells in Figure 3.11 show the optimal solution obtained after running Solver.
This solution is
Assign Ann to prepare conference packets.
Assign Ian to do the computer graphics.
Assign Joan to handle registrations.
Assign Sean to do the word processing.
The total cost given in cell J30 is $1,957.
Characteristics of Assignment Problems
Note that all the functional constraints of the Sellmore Co. problem (as shown in cells H24:J27
and D28:G30 of Figure 3.11) are fixed-requirement constraints which require each person to
perform exactly one task and require each task to be performed by exactly one person. Thus,
like the Big M Company transportation problem, the Sellmore Co. is a fixed-requirements
problem. This is a characteristic of all pure assignment problems. However, Chapter 15 (available at www.mhhe.com/Hillier6e) gives some examples of variants of assignment problems
where this is not the case.
Like the changing cells Assignment (D24:G27) in Figure 3.11, the changing cells in
the spreadsheet model for any pure assignment problem gives a value of 1 when the corresponding assignment is being made and a value of 0 otherwise. Since the fixed-requirement
constraints require only each row or column of changing cells to add up to 1 (which could
happen, e.g., if two of the changing cells in the same row or column had a value of 0.5 and
the rest 0), this would seem to necessitate adding the constraints that each of the changing
cells must be integer. After choosing the Solver option to make the changing cells nonnegative, this then would force each of the changing cells to be 0 or 1. However, it turned out to
be unnecessary to add the constraints that require the changing cells to have values of 0 or 1
3.6 Assignment Problems 101
FIGURE 3.11
A spreadsheet formulation of the Sellmore Co. problem as an assignment problem, including the objective cell TotalCost (J30) and
the other output cells Cost (D15:G18), TotalAssignments (H24:H27), and TotalAssigned (D28:G28), as well as the specifications
needed to set up the model. The values of 1 in the changing cells Assignment (D24:G27) show the optimal plan obtained by Solver
for assigning the people to the tasks.
A
1
2
B
C
D
E
F
G
H
I
J
Sellmore Co. Assignment Problem
Task
3
4
Required Time
5
(Hours)
Word
Hourly
Processing
Graphics
Packets
Registrations
Wage
Ann
35
41
27
40
$14
Ian
47
45
32
51
$12
8
Joan
39
56
36
43
$13
9
Sean
32
51
25
46
$15
6
Assignee
7
10
11
Task
12
Word
13
Processing
Graphics
Packets
Registrations
Ann
$490
$574
$378
$560
Ian
$564
$540
$384
$612
17
Joan
$507
$728
$468
$559
18
Sean
$480
$765
$375
$690
Cost
14
15
16
Assignee
19
20
Task
21
Word
Assignment
22
Total
Processing
Graphics
Packets
Registrations
Assignments
Ann
0
0
1
0
1
=
1
Ian
0
1
0
0
1
=
1
26
Joan
0
0
0
1
1
=
1
27
Sean
1
0
0
0
1
=
1
28
Total Assigned
1
1
1
1
=
=
=
1
1
1
=
1
23
24
25
Assignee
29
Demand
30
B
C
D
Cost
15
Ann
F
Total Cost
$1,957
H
G
Word
22
Total
Processing
23
Assignments
=G6*I6
24
=SUM(D24:G24)
=SUM(D25:G25)
13
14
E
Supply
=D6*I6
Graphics
Packets
Registrations
=F6*I6
=E6*I6
Ian
=D7*I7
=E7*I7
=F7*I7
=G7*I7
25
17
Joan
=D8*I8
=E8*I8
=F8*I8
=G8*I8
26
=SUM(D26:G26)
18
Sean
=D9*I9
=E9*I9
=F9*I9
=G9*I9
27
=SUM(D27:G27)
16
Assignee
Solver Parameters
Set Objective Cell: TotalCost
To: Min
By Changing Variable Cells:
Assignment
Subject to the Constraints:
TotalAssigned = Demand
TotalAssignments = Supply
J
29
Total Cost
30
=SUMPRODUCT(Cost,Assignment)
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
28
C
D
E
Total Assigned
=SUM(D24:D27)
=SUM(E24:E27)
F
G
=SUM(F24:F27) =SUM(G24:G27)
Range Name
Cells
Assignment
Cost
Demand
HourlyWage
RequiredTime
Supply
TotalAssigned
TotalAssignments
TotalCost
D24:G27
D15:G18
D30:G30
I6:I9
D6:G9
J24:J27
D28:G28
H24:H27
J30
An Application Vignette
Taylor Communications (formerly the Standard Operations
Company before its acquisition and rebranding by the Taylor
Corporation in 2016) is one of the largest printing conglomerates in the world. It provides high volume print production of
forms and print stationery for major firms in and near the United
States. It operates dozens of manufacturing facilities across the
United States and Mexico.
The print market is highly competitive, so company management must continually address the strategic challenge of minimizing the total cost of producing and distributing its products in order
to offer competitive pricing. A wide variety of printing presses are
available in the various manufacturing facilities and each press has
the capacity to perform multiple print jobs. However, the cost of
performing any given job can differ greatly from one press to the
next. Furthermore, the cost of transporting the finished product to
the customer can vary greatly depending on which manufacturing
facility is used and the resulting distance to the customer.
Given a large number of print jobs that need to be performed
over the next time period, the problem is to assign the print jobs
to presses. The cost of each individual assignment is the sum of
the cost of producing that job on that press and the cost of transporting the finished job to the customer. The objective is to select
all of the assignments of jobs to presses so as to minimize the total
cost of all the assignments. This is in fact an assignment problem
as described in the current section except for the modification
that a given press might be able to handle multiple jobs. Therefore, a constraint is added for each press that the total processing time for all of the jobs being assigned to that press cannot
exceed the total processing time that is available at that press.
Periodically solving this very large assignment problem and
implementing this solution has provided an estimated annual
savings of over $10 million.
Source: S. L. Ahire, M. F. Gorman, D. Dwiggins, and O. Mudry, “Operations Research Helps Reshape Operations Strategy at Standard
Register Company,” Interfaces 37, no. 6 (November-December
2007), pp.553–565. (A link to this article is available at
www.mhhe.com/Hillier6e.)
in Figure 3.11 because Solver gave an optimal solution that had only values of 0 or 1 anyway.
In fact, a general characteristic of pure assignment problems is that Solver always provides
such an optimal solution without needing to add these additional constraints.
As described further in Chapter 15, available at www.mhhe.com/Hillier6e, another interesting characteristic of any pure assignment problem is that it can be viewed as a special type
of pure transportation problem. In particular, every fixed-requirement constraint in the corresponding transportation problem would require that either a row or column of changing cells
add up to 1. This would result in Solver giving an optimal solution where every changing cell
has a value of either 0 or 1, just as for the original assignment problem.
Review
Questions
3.7
1. Why are assignment problems given this name?
2. Pure assignment problems have what type of functional constraints?
3. What is the interpretation of the changing cells in the spreadsheet model of a pure assignment problem?
MODEL FORMULATION FROM A BROADER PERSPECTIVE
Formulating and analyzing a linear programming model provides information to help managers make their decisions. That means the model must accurately reflect the managerial view
of the problem:
Both the measure of performance and the constraints
in a model need to reflect
the managerial view of the
problem.
102
• The overall measure of performance must capture what management wants accomplished.
• When management limits the amounts of resources that will be made available to the activities under consideration, these limitations should be expressed as resource constraints.
• When management establishes minimum acceptable levels for benefits to be gained from
the activities, these managerial goals should be incorporated into the model as benefit
constraints.
• If management has fixed requirements for certain quantities, then fixed-requirement constraints are needed.
With the help of spreadsheets, some managers now are able to formulate and solve small
linear programming models themselves. However, larger linear programming models may be
formulated by management science teams, not managers. When this is done, the management
3.7 Model Formulation from a Broader Perspective 103
Linear programming studies need strong managerial
input and support.
What-if analysis addresses
some key questions that
remain after formulating
and solving a model.
science team must thoroughly understand the managerial view of the problem. This requires clear
communication with management from the very beginning of the study and maintaining effective
communication as new issues requiring managerial guidance are identified. Management needs
to clearly convey its view of the problem and the important issues involved. A manager cannot
expect to obtain a helpful linear programming study without making clear just what help is
wanted.
As is necessary in any textbook, the examples in this chapter are far smaller, simpler, and
more clearly spelled out than is typical of real applications. Many real studies require formulating complicated linear programming models involving hundreds or thousands (or possibly
even millions) of decisions and constraints. In these cases, there usually are many ambiguities
about just what should be incorporated into the model. Strong managerial input and support
are vital to the success of a linear programming study for such complex problems.
When dealing with huge real problems, there is no such thing as “the” correct linear programming model for the problem. The model continually evolves throughout the course of the
study. Early in the study, various techniques are used to test initial versions of the model to
identify the errors and omissions that inevitably occur when constructing such a large model.
This testing process is referred to as model validation.
Once the basic formulation has been validated, there are many reasonable variations of the
model that might be used. Which variation to use depends on such factors as the assumptions
about the problem that seem most reasonable, the estimates of the parameters of the model
that seem most reliable, and the degree of detail desired in the model.
In large linear programming studies, a good approach is to begin with a relatively simple
version of the model and then use the experience gained with this model to evolve toward
more elaborate models that more nearly reflect the complexity of the real problem. This process of model enrichment continues only as long as the model remains reasonably easy to
solve. It must be curtailed when the study’s results are needed by management. Managers
often need to curb the natural instinct of management science teams to continue adding “bells
and whistles” to the model rather than winding up the study in a timely fashion with a less
elegant but adequate model.
When managers study the output of the current model, they often detect some undesirable
characteristics that point toward needed model enrichments. These enrichments frequently
take the form of new benefit constraints to satisfy some managerial goals not previously articulated. (Recall that this is what happened in the Super Grain case study.)
Even though many reasonable variations of the model could be used, an optimal solution can
be solved for only with respect to one specific version of the model at a time. This is why whatif analysis is such an important part of a linear programming study. After obtaining an optimal
solution with respect to one specific model, management will have many what-if questions:
• What if the estimates of the parameters in the model are incorrect?
• How do the conclusions change if different plausible assumptions are made about the problem?
• What happens when certain managerial options are pursued that are not incorporated into
the current model?
Chapter 5 is devoted primarily to describing how what-if analysis addresses these and related
issues, as well as how managers use this information.
Because managers instigate management science studies, they need to know enough about
linear programming models and their formulation to be able to recognize managerial problems to
which linear programming can be applied. Furthermore, since managerial input is so important for linear programming studies, managers need to understand the kinds of managerial
concerns that can be incorporated into the model. Developing these two skills have been the
most important goals of this chapter.
Review
Questions
1.
2.
3.
4.
A linear programming model needs to reflect accurately whose view of the problem?
What is meant by model validation?
What is meant by the process of model enrichment?
Why is what-if analysis an important part of a linear programming study?
104 Chapter Three Linear Programming: Formulation and Applications
3.8 Summary
Functional constraints with a ≤ sign are called resource constraints, because they require that the amount
used of some resource must be less than or equal to the amount available of that resource. The identifying feature of resource-allocation problems is that all their functional constraints are resource constraints.
Functional constraints with a ≥ sign are called benefit constraints, since their form is that the level
achieved for some benefit must be greater than or equal to the minimum acceptable level for that benefit. Frequently, benefit constraints express goals prescribed by management. If every functional constraint is a benefit constraint, then the problem is a cost–benefit–trade-off problem.
Functional constraints with an = sign are called fixed-requirement constraints, because they express the
fixed requirement that, for some quantity, the amount provided must be equal to the required amount. The
identifying feature of fixed-requirements problems is that their functional constraints are fixed-requirement
constraints. One prominent type of fixed-requirements problem is transportation problems, which typically
involve finding a shipping plan that minimizes the total cost of transporting a product from a number of
plants to a number of customers. Another prominent type is assignment problems, which typically involves
assigning people to tasks so as to minimize the total cost of performing these tasks.
Linear programming problems that do not fit into any of these three categories are called mixed problems.
In many real applications, management science teams formulate and analyze large linear programming models to help guide managerial decision making. Such teams need strong managerial input and
support to help ensure that their work really meets management’s needs.
Glossary
assignment problem A type of linear programming problem that typically involves assigning
people to tasks so as to minimize the total cost of
performing these tasks. (Section 3.6), 99
benefit constraint A functional constraint
with a ≥ sign. The left-hand side is interpreted as
the level of some benefit that is achieved by the
activities under consideration, and the right-hand
side is the minimum acceptable level for that benefit. (Section 3.3), 82
cost–benefit–trade-off problem A type of
linear programming problem involving the
trade-off between the total cost of the activities under consideration and the benefits to be
achieved by these activities. Its identifying feature is that each functional constraint in the linear programming model is a benefit constraint.
(Section 3.3), 82
fixed-requirement constraint A functional
constraint with an = sign. The left-hand side represents the amount provided of some type of quantity, and the right-hand side represents the required
amount for that quantity. (Section 3.4), 88
fixed-requirements problem A type of linear
programming problem concerned with optimizing how to meet a number of fixed requirements.
Its identifying feature is that each functional
constraint in its model is a fixed-requirement constraint. (Section 3.4), 88
identifying feature A feature of a model that
identifies the category of linear programming
problem it represents. (Chapter introduction), 64
integer programming problem A variation of a
linear programming problem that has the additional
restriction that some or all of the decision variables
must have integer values. (Section 3.2), 75
mixed problem Any linear programming problem that includes at least two of the three types
of functional constraints (resource constraints,
benefit constraints, and fixed-requirement constraints). (Section 3.4), 88
model enrichment The process of using experience with a model to identify and add important
details that will provide a better representation of
the real problem. (Section 3.7), 103
model validation The process of checking and
testing a model to develop a valid model. (Section
3.7), 103
resource-allocation problem A type of linear
programming problem concerned with allocating
resources to activities. Its identifying feature is
that each functional constraint in its model is a
resource constraint. (Section 3.2), 71
resource constraint A functional constraint
with a ≤ sign. The left-hand side represents the
amount of some resource that is used by the
activities under consideration, and the righthand side represents the amount available of that
resource. (Section 3.2), 71
transportation problem A type of linear programming problem that typically involves finding
a shipping plan that minimizes the total cost of
transporting a product from a number of plants to
a number of customers. (Section 3.5), 95
Learning Aids for This Chapter
All learning aids are available at www.mhhe.com/Hillier6e.
Excel Files:
Super Grain Example
TBA Airlines Example
Think-Big Example
Union Airways Example
Big M Example
Revised Super Grain Example
Sellmore Example
Excel Add-in:
Analytic Solver
Chapter 3 Solved Problems 105
Solved Problems
The solutions are available at www.mhhe.com/Hillier6e.
3.S1. Farm Management
Dwight and Hattie have run the family farm for over 30 years.
They are currently planning the mix of crops to plant on their
120-acre farm for the upcoming season. The table gives the
labor-hours and fertilizer required per acre, as well as the total
expected profit per acre for each of the potential crops under
consideration. Dwight, Hattie, and their children can work at
most 6,500 total hours during the upcoming season. They have
200 tons of fertilizer available. What mix of crops should be
planted to maximize the family’s total profit?
a. Formulate and solve a linear programming model for this
problem in a spreadsheet.
b. Formulate this same model algebraically.
Crop
Labor
Required
(hours per
acre)
Fertilizer
Required
(tons per
acre)
Expected
Profit (per
acre)
Oats
Wheat
Corn
50
60
105
1.5
2
4
$500
$600
$950
12-inch rods (with no waste). For the next production period,
Decora needs twenty-five 12-inch rods, fifty-two 18-inch rods,
forty-five 24-inch rods, thirty 40-inch rods, and twelve 60-inch
rods. What is the fewest number of 60-inch rods that can be
purchased to meet their production needs? Formulate and solve
an integer programming model in a spreadsheet.
3.S4. Producing and Distributing AEDs at Heart Start
Heart Start produces automated external defibrillators in each
of two different plants (A and B). The unit production costs
and monthly production capacity of the two plants are indicated in the table below. The automated external defibrillators
are sold through three wholesalers. The shipping cost from
each plant to the warehouse of each wholesaler along with the
monthly demand from each wholesaler are also indicated in
the table. The management of Heart Start now has asked their
top management scientist (you) to address the following two
questions. How many automated external defibrillators should
be produced in each plant, and how should they be distributed
to each of the three wholesaler warehouses so as to minimize
the combined cost of production and shipping? Formulate and
solve a linear programming model in a spreadsheet.
Unit Shipping Cost
3.S2. Diet Problem
The kitchen manager for Sing Sing prison is trying to decide
what to feed its prisoners. She would like to offer some combination of milk, beans, and oranges. The goal is to minimize
cost, subject to meeting the minimum nutritional requirements
imposed by law. The cost and nutritional content of each food,
along with the minimum nutritional requirements, are shown
below. What diet should be fed to each prisoner?
a. Formulate and solve a linear programming model for this
problem in a spreadsheet.
b. Formulate this same model algebraically.
Oranges
Navy
(large
Minimum
Milk
Beans
Calif.
Daily
(gallons) (cups) Valencia) Requirement
Niacin (mg)
Thiamin (mg)
Vitamin C
(mg)
Cost ($)
3.2
1.12
32.0
2.00
4.9
1.3
0.0
0.20
3.S3. Cutting Stock Problem
0.8
0.19
13.0
1.5
93.0
0.25
45.0
Decora Accessories manufactures a variety of bathroom accessories, including decorative towel rods and shower curtain
rods. Each of the accessories includes a rod made out of stainless steel. However, many different lengths are needed: 12, 18,
24, 40, and 60 inches. Decora purchases 60-inch rods from
an outside supplier and then cuts the rods as needed for their
products. Each 60-inch rod can be used to make a number of
smaller rods. For example, a 60-inch rod could be used to make
a 40-inch and an 18-inch rod (with 2 inches of waste), or five
Plant A
Plant B
Monthly
Demand
Unit Monthly
Ware- Produc- Produchouse
tion
tion
3
Cost Capacity
Warehouse
1
Warehouse
2
$22
$16
$14
$20
$30
$24
80
60
70
$600
$625
100
120
3.S5. Bidding for Classes
In the MBA program at a prestigious university in the Pacific
Northwest, students bid for electives in the second year of their
program. Each student has 100 points to bid (total) and must take
two electives. There are four electives available: Management
Science (MS), Finance (Fin), Operations Management (OM), and
Marketing (Mkt). Each class is limited to 5 students. The bids submitted for each of the 10 students are shown in the table below.
Student Bids for Classes
Student
MS
Fin
OM
Mkt
George
Fred
Ann
Eric
Susan
Liz
Ed
David
Tony
Jennifer
60
20
45
50
30
50
70
25
35
60
10
20
45
20
30
50
20
25
15
10
10
40
5
5
30
0
10
35
35
10
20
20
5
25
10
0
0
15
15
20
106 Chapter Three Linear Programming: Formulation and Applications
a. Formulate and solve a spreadsheet model to determine an assignment of students to classes so as to maximize the total bid
points of the assignments.
b. Does the resulting solution seem like a fair assignment?
c. Which alternative objectives might lead to a fairer assignment?
Problems
An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.
3.1.
Reconsider the Super Grain Corp. case study as presented in Section 3.1. The advertising firm, Giacomi & Jackowitz,
now has suggested a fourth promising advertising medium—radio
commercials—to promote the company’s new breakfast cereal,
Crunchy Start. Young children are potentially major consumers of
this cereal, but parents of young children (the major potential purchasers) often are too busy to do much reading (so may miss the
company’s advertisements in magazines and Sunday supplements)
or even to watch the Saturday morning programs for children
where the company’s television commercials are aired. However,
these parents do tend to listen to the radio during the commute to
and from work. Therefore, to better reach these parents, Giacomi
& Jackowitz suggests giving consideration to running commercials for Crunchy Start on nationally syndicated radio programs
that appeal to young adults during typical commuting hours.
Giacomi & Jackowitz estimates that the cost of developing each
new radio commercial would be $50,000, and that the expected
number of exposures per commercial would be 900,000. The firm
has determined that 10 spots are available for different radio commercials, and each one would cost $200,000 for a normal run.
a. Formulate and solve a spreadsheet model for the
revised advertising-mix problem that includes this
fourth advertising medium. Identify the data cells,
the changing cells, and the objective cell. Also show
the Excel equation for each output cell expressed as
a SUMPRODUCT function.
b. Indicate why this spreadsheet model is a linear programming model.
c. Express this model in algebraic form.
3.2.
Read the referenced article that fully describes the management science study summarized in the application vignette
presented in Section 3.2. Briefly describe how linear programming was applied in this study. Then list the various benefits that
resulted from this study.
3.3.*
Consider a resource-allocation problem having the following data.
Resource Usage
per Unit of
Each Activity
Resource
1
2
3
Contribution
per unit
1
2
Amount of
Resource Available
2
3
2
1
3
4
10
20
20
$20
$30
Contribution per unit = profit per unit of the activity.
a. Formulate a linear programming model for this
problem on a spreadsheet.
b. Use the spreadsheet to check the following solutions: (x1, x2) = (2, 2), (3, 3), (2, 4), (4, 2), (3, 4),
(4, 3). Which of these solutions are feasible? Which
of these feasible solutions has the best value of the
objective function?
c. Use Solver to find an optimal solution.
d. Express this model in algebraic form.
e. Use the graphical method to solve this model.
3.4.
Consider a resource-allocation problem having the following data.
Resource Usage
per Unit of
Each Activity
Resource
1
2
3
Amount of
Resource
Available
A
B
C
30
0
20
20
10
20
0
40
30
500
600
1,000
$50
$40
$70
Contribution per unit
Contribution per unit = profit per unit of the activity.
a. Formulate and solve a linear programming model
for this problem on a spreadsheet.
b. Express this model in algebraic form.
3.5.
Consider a resource-allocation problem having the following data.
Resource Usage per
Unit of Each Activity
Resource
1
2
3
4
Amount of
Resource
Available
P
Q
R
S
3
4
6
−2
5
−1
3
2
−2
3
2
5
4
2
−1
3
400
300
400
300
$11
$9
$8
$9
Contribution
per unit
Contribution per unit = profit per unit of the activity.
a. Formulate a linear programming model for this
problem on a spreadsheet.
b. Make five guesses of your own choosing for the
optimal solution. Use the spreadsheet to check each
one for feasibility and, if feasible, for the value of
the objective function. Which feasible guess has the
best objective function value?
c. Use Solver to find an optimal solution.
Chapter 3 Problems 107
3.6.*
The Omega Manufacturing Company has discontinued
the production of a certain unprofitable product line. This act
created considerable excess production capacity. Management
is considering devoting this excess capacity to one or more of
three products, Products 1, 2, and 3. The available capacity of
the machines that might limit output is summarized in the following table.
Machine Type
Available Time
(in Machine-Hours per Week)
Milling machine
Lathe
Grinder
500
350
150
The number of machine-hours required for each unit of the
respective products are shown in the next table.
Productivity Coefficient (in Machine-Hours per Unit)
Machine Type
Product 1
Product 2
Product 3
Milling machine
Lathe
Grinder
9
5
3
3
4
0
5
0
2
Each machine is available 40 hours per month. Each part
manufactured will yield a unit profit as follows:
Part
Profit
a. Indicate why this is a resource-allocation problem
by identifying both the activities and the limited
resources to be allocated to these activities.
b. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure
of performance for the decisions.
c. Convert these verbal descriptions of the constraints
and the measure of performance into quantitative
expressions in terms of the data and decisions.
d. Formulate a spreadsheet model for this problem. Identify the data cells, the changing cells, the objective cell,
and the other output cells. Also show the Excel equation for each output cell expressed as a SUMPRODUCT function. Then use Solver to solve the model.
e. Summarize the model in algebraic form.
3.7.
Ed Butler is the production manager for the Bilco Corporation, which produces three types of spare parts for automobiles. The manufacture of each part requires processing on
each of two machines, with the following processing times (in
hours).
Part
Machine
1
2
A
B
C
0.02
0.05
0.03
0.02
0.05
0.04
B
C
$50
$40
$30
Ed wants to determine the mix of spare parts to produce to
maximize total profit.
a. Identify both the activities and the resources for this
resource-allocation problem.
b. Formulate a linear programming model for this
problem on a spreadsheet.
c. Make three guesses of your own choosing for the
optimal solution. Use the spreadsheet to check each
one for feasibility and, if feasible, for the value of
the objective function. Which feasible guess has the
best objective function value?
d. Use Solver to find an optimal solution.
e. Express the model in algebraic form.
3.8.
Consider the following algebraic formulation of a
resource-allocation problem with three resources, where the decisions to be made are the levels of three activities (A1, A2, and A3).
Maximize
The Sales Department indicates that the sales potential for
Products 1 and 2 exceeds the maximum production rate and that
the sales potential for product 3 is 20 units per week. The unit
profit would be $50, $20, and $25, respectively, for Products 1,
2, and 3. The objective is to determine how much of each product Omega should produce to maximize profit.
A
Profit = 20 A1 + 40 A2 + 30 A3
subject to
Resource 1: 3 A1 + 5 A2 + 4 A3 ≤ 400 (amount available)
Resource 2: A1 + A2 + A3 ≤ 100 (amount available)
Resource 3: A1 + 3 A2 + 2 A3 ≤ 200 (amount available)
and
A1 ≥ 0 A2 ≥ 0 A3 ≥ 0
Formulate and solve the spreadsheet model for this problem.
3.9.
Consider a cost–benefit–trade-off problem having the
following data.
Benefit Contribution per Unit of Each
Activity
Benefit
1
2
1
2
3
Unit cost
5
2
7
$60
3
2
9
$50
Minimum
Acceptable Level
60
30
126
a. Formulate a linear programming model for this
problem on a spreadsheet.
b. Use the spreadsheet to check the following solutions:
(x1, x2) = (7, 7), (7, 8), (8, 7), (8, 8), (8, 9), (9, 8). Which
of these solutions are feasible? Which of these feasible
solutions has the best value of the objective function?
c. Use Solver to find an optimal solution.
d. Express the model in algebraic form.
e. Use the graphical method to solve this model.
108 Chapter Three Linear Programming: Formulation and Applications
3.10.
Consider a cost–benefit–trade-off problem having the
following data.
Benefit Contribution
per Unit of Each Activity
Benefit
1
2
3
4
Minimum
Acceptable
Level
P
Q
R
2
1
3
−1
4
5
4
−1
4
3
2
−1
80
60
110
Unit cost
$400
$600
$500
$300
a. Formulate a linear programming model for this
problem on a spreadsheet.
b. Make five guesses of your own choosing for the
optimal solution. Use the spreadsheet to check each
one for feasibility and, if feasible, for the value of
the objective function. Which feasible guess has the
best objective function value?
c. Use Solver to find an optimal solution.
3.11.* Fred Jonasson manages a family-owned farm. To supplement several food products grown on the farm, Fred also raises
pigs for market. He now wishes to determine the quantities of the
available types of feed (corn, tankage, and alfalfa) that should be
given to each pig. Since pigs will eat any mix of these feed types,
the objective is to determine which mix will meet certain nutritional requirements at a minimum cost. The number of units of
each type of basic nutritional ingredient contained within a kilogram of each feed type is given in the following table, along with
the daily nutritional requirements and feed costs.
construction costs, Maureen needs to invest some of the company’s money now to meet these future cash flow needs. Maureen
may purchase only three kinds of financial assets, each of which
costs $1 million per unit. Fractional units may be purchased. The
assets produce income 5, 10, and 20 years from now, and that
income is needed to cover minimum cash flow requirements in
those years, as shown in the following table.
Income per Unit of Asset
Year
Asset 1
Asset 2
Asset 3
Minimum
Cash Flow
Required
5
10
20
$8 million
2 million
0
$4 million
2 million
6 million
$2 million
4 million
8 million
$1.6 billion
400 million
1.2 billion
Maureen wishes to determine the mix of investments in these
assets that will cover the cash flow requirements while minimizing the total amount invested.
a. Formulate a linear programming model for this
problem on a spreadsheet.
b. Use the spreadsheet to check the possibility of purchasing 100 units of asset 1, 100 units of asset 2, and
200 units of asset 3. How much cash flow would this
mix of investments generate 5, 10, and 20 years from
now? What would be the total amount invested?
c. Take a few minutes to use a trial-and-error approach
with the spreadsheet to develop your best guess
for the optimal solution. What is the total amount
invested for your solution?
Kilogram
of Corn
Kilogram of
Tankage
Kilogram
of Alfalfa
Minimum
Daily
Requirement
Carbohydrates
Protein
Vitamins
90
30
10
20
80
20
40
60
60
200
180
150
Cost (¢)
84
72
60
Nutritional
Ingredient
a. Formulate a linear programming model for this
problem on a spreadsheet.
b. Use the spreadsheet to check if (x1, x2, x3) = (1, 2, 2)
is a feasible solution and, if so, what the daily cost
would be for this diet. How many units of each
nutritional ingredient would this diet provide daily?
c. Take a few minutes to use a trial-and-error approach
with the spreadsheet to develop your best guess for
the optimal solution. What is the daily cost for your
solution?
d. Use Solver to find an optimal solution.
e. Express the model in algebraic form.
3.12.
Maureen Laird is the chief financial officer for the Alva
Electric Co., a major public utility in the Midwest. The company has scheduled the construction of new hydroelectric plants
5, 10, and 20 years from now to meet the needs of the growing
population in the region served by the company. To cover the
d. Use Solver to find an optimal solution.
e. Summarize the model in algebraic form.
3.13. Web Mercantile sells many household products
through an online catalog. The company needs substantial
warehouse space for storing its goods. Plans now are being
made for leasing warehouse storage space over the next five
months. Just how much space will be required in each of these
months is known. However, since these space requirements are
quite different, it may be most economical to lease only the
amount needed each month on a month-by-month basis. On the
other hand, the additional cost for leasing space for additional
months is much less than for the first month, so it may be less
expensive to lease the maximum amount needed for the entire
five months. Another option is the intermediate approach of
changing the total amount of space leased (by adding a new
lease and/or having an old lease expire) at least once but not
every month.
Chapter 3 Problems 109
The space requirement and the leasing costs for the various
leasing periods are as follows.
Month
Required Space
(Square Feet)
1
2
3
4
5
30,000
20,000
40,000
10,000
50,000
Leasing Period
(Months)
Cost per
Sq. Ft. Leased
1
2
3
4
5
$ 65
100
135
160
190
The objective is to minimize the total leasing cost for meeting the space requirements.
a. Indicate why this is a cost–benefit–trade-off problem by identifying both the activities and the benefits being sought from these activities.
b. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure
of performance for the decisions.
c. Convert these verbal descriptions of the constraints
and the measure of performance into quantitative
expressions in terms of the data and decisions.
d. Formulate a spreadsheet model for this problem. Identify the data cells, the changing cells, the objective cell,
and the other output cells. Also show the Excel equation for each output cell expressed as a SUMPRODUCT function. Then use Solver to solve the model.
e. Summarize the model in algebraic form.
3.14.
Consider the following algebraic formulation of a cost–
benefit–trade-off problem involving three benefits, where the
decisions to be made are the levels of four activities (A1, A2, A3,
and A4):
Minimize Cost = 2 A1 + A2 − A3 + 3 A4
subject to
Benefit 1: 3 A1 + 2 A2 − 2 A3 + 5 A4 ≥ 80 (minimum
acceptable level)
Benefit 2: A1 − A2
+ A4 ≥ 10 (minimum
acceptable level)
Benefit 3: A1 + A2 −
and
A3 + 2 A4 ≥ 30 (minimum
acceptable
level)
A1 ≥ 0 A2 ≥ 0 A3 ≥ 0 A4 ≥ 0
Formulate and solve the spreadsheet model for this problem.
3.15.
Larry Edison is the director of the Computer Center for
Buckly College. He now needs to schedule the staffing of the center. It is open from 8 AM until midnight. Larry has monitored the
usage of the center at various times of the day and determined that
the following number of computer consultants (qualified graduate
students) are required.
Minimum Number of
Consultants Required
to Be on Duty
Time of Day
8 AM–noon
Noon–4 PM
4 PM–8 PM
8 PM–midnight
6
8
12
6
Two types of computer consultants can be hired: full-time
and part-time. The full-time consultants work for eight consecutive hours in any of the following shifts: morning (8 AM–4 PM),
afternoon (noon–8 PM), and evening (4 PM–midnight). Full-time
consultants are paid $17.50 per hour.
Part-time consultants can be hired to work any of the four shifts
listed in the table. Part-time consultants are paid $15 per hour.
An additional requirement is that during every time period,
there must be at least two full-time consultants on duty for every
part-time consultant on duty.
Larry would like to determine how many full-time and
part-time consultants should work each shift to meet the above
requirements at the minimum possible cost.
a. Which category of linear programming problem
does this problem fit? Why?
b. Formulate and solve a linear programming model
for this problem on a spreadsheet.
c. Summarize the model in algebraic form.
3.16.* The Medequip Company produces precision medical
diagnostic equipment at two factories. Three medical centers
have placed orders for this month’s production output. The following table shows what the cost would be for shipping each unit
from each factory to each of these customers. Also shown are
the number of units that will be produced at each factory and the
number of units ordered by each customer.
A decision now needs to be made about the shipping plan
for how many units to ship from each factory to each customer.
a. Which category of linear programming problem
does this problem fit? Why?
b. Formulate and solve a linear programming model
for this problem on a spreadsheet.
c. Summarize this formulation in algebraic form.
Unit Shipping Cost
From
To
Customer Customer Customer
1
2
3
Output
Factory 1
$600
$800
$700
Factory 2
400
900
600
300 units
200 units
Order size
400 units
400
units
500
units
110 Chapter Three Linear Programming: Formulation and Applications
3.17.
Read the referenced article that fully describes the management science study summarized in the application vignette
presented in Section 3.4. Briefly describe how linear programming was applied in this study. Then list the various benefits that
resulted from this study.
3.18.
The Fagersta Steelworks currently is working two mines
to obtain its iron ore. This iron ore is shipped to either of two
storage facilities. When needed, it then is shipped on to the company’s steel plant. The diagram below depicts this distribution
network, where M1 and M2 are the two mines, S1 and S2 are the
two storage facilities, and P is the steel plant. The diagram also
shows the monthly amounts produced at the mines and needed at
the plant, as well as the shipping cost and the maximum amount
that can be shipped per month through each shipping lane.
40 tons
produced
M1
$2,000/ton
30 tons max.
S1
n
/to x.
00 ma
,7
$1 ons
t
30
70
M2
$1,100/ton
50 tons max.
100 tons
needed
P
$
50 1,60
to 0/t
ns
o
m n
ax
.
60 tons
produced
$4
00
ton /ton
sm
ax
.
S2
n
/to x.
00
$8 s ma
ton
70
Management now wants to determine the most economical
plan for shipping the iron ore from the mines through the distribution network to the steel plant.
a. Identify all the requirements that will need to be
expressed in fixed-requirement constraints.
b. Formulate and solve a linear programming model
for this problem on a spreadsheet.
c. Express this model in algebraic form.
3.19.* Al Ferris has $60,000 that he wishes to invest now in
order to use the accumulation for purchasing a retirement annuity in five years. After consulting with his financial advisor, he
has been offered four types of fixed-income investments, which
we will label as investments A, B, C, and D.
Investments A and B are available at the beginning of each of the
next five years (call them years 1 to 5). Each dollar invested in A at
the beginning of a year returns $1.40 (a profit of $0.40) two years
later (in time for immediate reinvestment). Each dollar invested in B
at the beginning of a year returns $1.70 three years later.
Investments C and D will each be available at one time in
the future. Each dollar invested in C at the beginning of year 2
returns $1.90 at the end of year 5. Each dollar invested in D at
the beginning of year 5 returns $1.30 at the end of year 5.
Al wishes to know which investment plan maximizes the amount
of money that can be accumulated by the beginning of year 6.
a. For this problem, all its functional constraints can
be expressed as fixed-requirement constraints. To
do this, let At, Bt, Ct, and Dt be the amounts invested
in investments A, B, C, and D, respectively, at the
beginning of year t for each t where the investment
is available and will mature by the end of year 5.
Also let Rt be the number of available dollars not
invested at the beginning of year t (and so available
for investment in a later year). Thus, the amount
invested at the beginning of year t plus Rt must
equal the number of dollars available for investment
at that time. Write such an equation in terms of the
relevant variables above for the beginning of each
of the five years to obtain the five fixed-requirement
constraints for this problem.
b. Formulate a complete linear programming model
for this problem in algebraic form.
c. Formulate and solve this model on a spreadsheet.
3.20.
The Metalco Company desires to blend a new alloy of
40 percent tin, 35 percent zinc, and 25 percent lead from several
available alloys having the following properties.
Alloy
Property
1
2
3
4
5
Percentage of tin
Percentage of zinc
Percentage of lead
Cost ($/lb)
60
10
30
22
25
15
60
20
45
45
10
25
20
50
30
24
50
40
10
27
The objective is to determine the proportions of these alloys
that should be blended to produce the new alloy at a minimum cost.
a. Identify all the requirements that will need to be
expressed in fixed-requirement constraints.
b. Formulate and solve a linear programming model
for this problem on a spreadsheet.
c. Express this model in algebraic form.
3.21. The Weigelt Corporation has three branch plants with
excess production capacity. Fortunately, the corporation has
a new product ready to begin production, and all three plants
have this capability, so some of the excess capacity can be
used in this way. This product can be made in three sizes—
large, medium, and small—that yield a net unit profit of $420,
$360, and $300, respectively. Plants 1, 2, and 3 have the excess
capacity to produce 750, 900, and 450 units per day of this
product, respectively, regardless of the size or combination of
sizes involved.
The amount of available in-process storage space also
imposes a limitation on the production rates of the new product.
Plants 1, 2, and 3 have 13,000, 12,000, and 5,000 square feet,
respectively, of in-process storage space available for a day’s
production of this product. Each unit of the large, medium, and
small sizes produced per day requires 20, 15, and 12 square feet,
respectively.
Sales forecasts indicate that if available, 900, 1,200, and 750
units of the large, medium, and small sizes, respectively, would
be sold per day.
At each plant, some employees will need to be laid off unless
most of the plant’s excess production capacity can be used to
produce the new product. To avoid layoffs if possible, management has decided that the plants should use the same percentage
of their excess capacity to produce the new product.
Management wishes to know how much of each of the sizes
should be produced by each of the plants to maximize profit.
a. Formulate and solve a linear programming model
for this mixed problem on a spreadsheet.
b. Express the model in algebraic form.
Chapter 3 Problems 111
3.22.* A cargo plane has three compartments for storing cargo:
front, center, and back. These compartments have capacity limits
on both weight and space, as summarized below.
Compartment
Front
Center
Back
Weight
Capacity
(Tons)
Space
Capacity
(Cubic Feet)
12
18
10
7,000
9,000
5,000
Furthermore, the weight of the cargo in the respective compartments must be the same proportion of that compartment’s
weight capacity to maintain the balance of the airplane.
The following four cargoes have been offered for shipment
on an upcoming flight as space is available.
Cargo
Weight
(Tons)
Volume
(Cubic Feet/Ton)
Profit
($/Ton)
1
2
3
4
20
16
25
13
500
700
600
400
320
400
360
290
Any portion of these cargoes can be accepted. The objective is to determine how much (if any) of each cargo should be
accepted and how to distribute each among the compartments to
maximize the total profit for the flight.
a. Formulate and solve a linear programming model
for this mixed problem on a spreadsheet.
b. Express the model in algebraic form.
3.23.
Comfortable Hands is a company that features a product line of winter gloves for the entire family—men, women, and
children. They are trying to decide what mix of these three types
of gloves to produce.
Comfortable Hands’s manufacturing labor force is unionized. Each full-time employee works a 40-hour week. In addition, by union contract, the number of full-time employees
can never drop below 20. Nonunion, part-time workers also
can be hired with the following union-imposed restrictions:
(1) each part-time worker works 20 hours per week and
(2) there must be at least two full-time employees for each parttime employee.
All three types of gloves are made out of the same 100
percent genuine cowhide leather. Comfortable Hands has a
long-term contract with a supplier of the leather and receives a
5,000-square-foot shipment of the material each week. The material requirements and labor requirements, along with the gross
profit per glove sold (not considering labor costs), are given in
the following table.
Glove
Men’s
Women’s
Children’s
Material
Required
(Square Feet)
Labor
Required
(Minutes)
Gross
Profit
(per Pair)
2
1.5
1
30
45
40
$8
10
6
Each full-time employee earns $13 per hour, while each
part-time employee earns $10 per hour. Management wishes to
know what mix of each of the three types of gloves to produce
per week, as well as how many full-time and part-time workers
to employ. They would like to maximize their net profit—their
gross profit from sales minus their labor costs.
a. Formulate and solve a linear programming model
for this problem on a spreadsheet.
b. Summarize this formulation in algebraic form.
3.24.
Oxbridge University maintains a powerful mainframe
computer for research use by its faculty, Ph.D. students, and
research associates. During all working hours, an operator must
be available to operate and maintain the computer, as well as to
perform some programming services. Beryl Ingram, the director
of the computer facility, oversees the operation.
It is now the beginning of the fall semester and Beryl is confronted with the problem of assigning different working hours to
her operators. Because all the operators are currently enrolled in
the university, they are available to work only a limited number
of hours each day.
There are six operators (four undergraduate students and two
graduate students). They all have different wage rates because of
differences in their experience with computers and in their programming ability. The following table shows their wage rates, along
with the maximum number of hours that each can work each day.
Maximum Hours of Availability
Operators Wage Rate Mon. Tue. Wed. Thurs. Fri.
K. C.
$20.00/hour
6
0
6
0
6
D. H.
$20.20/hour
0
6
0
6
0
H. B.
$19.80/hour
4
8
4
0
4
S. C.
$19.60/hour
5
5
5
0
5
K. S.
$21.60/hour
3
0
3
8
0
N. K.
$22.60/hour
0
0
0
6
2
Each operator is guaranteed a certain minimum number of
hours per week that will maintain an adequate knowledge of the
operation. This level is set arbitrarily at 8 hours per week for
the undergraduate students (K. C., D. H., H. B., and S. C.) and
7 hours per week for the graduate students (K. S. and N. K.).
The computer facility is to be open for operation from 8 AM
to 10 PM Monday through Friday with exactly one operator on
duty during these hours. On Saturdays and Sundays, the computer is to be operated by other staff.
112 Chapter Three Linear Programming: Formulation and Applications
12 grams (g) of protein. Furthermore, for practical reasons, each
child needs exactly 2 slices of bread (to make the sandwich), at
least twice as much peanut butter as jelly, and at least 1 cup of
liquid (milk and/or juice).
Joyce and Marvin would like to select the food choices
for each child that minimize cost while meeting the above
requirements.
Because of a tight budget, Beryl has to minimize cost. She
wishes to determine the number of hours she should assign to
each operator on each day. Formulate and solve a spreadsheet
model for this problem.
3.25.
Slim-Down Manufacturing makes a line of nutritionally
complete, weight-reduction beverages. One of its products is a
strawberry shake that is designed to be a complete meal. The
strawberry shake consists of several ingredients. Some information about each of these ingredients is given next.
Ingredient
Calories
from Fat
(per tbsp.)
Total Calories (per
tbsp.)
Vitamin
Content
(mg/tbsp.)
Thickeners
(mg/tbsp.)
Cost
(¢/tbsp.)
1
75
0
0
30
50
100
0
120
80
20
0
50
0
2
3
8
1
2
25
10
8
25
15
6
Strawberry flavoring
Cream
Vitamin supplement
Artificial sweetener
Thickening agent
The nutritional requirements are as follows. The beverage
must total between 380 and 420 calories (inclusive). No more
than 20 percent of the total calories should come from fat. There
must be at least 50 milligrams (mg) of vitamin content. For taste
reasons, there must be at least two tablespoons (tbsp.) of strawberry flavoring for each tbsp. of artificial sweetener. Finally, to
maintain proper thickness, there must be exactly 15 mg of thickeners in the beverage.
Management would like to select the quantity of each ingredient for the beverage that would minimize cost while meeting
the above requirements.
a. Identify the requirements that lead to resource constraints, to benefit constraints, and to fixed-requirement constraints.
b. Formulate and solve a linear programming model
for this problem on a spreadsheet.
c. Summarize this formulation in algebraic form.
3.26.
Joyce and Marvin run a day care for preschoolers. They
are trying to decide what to feed the children for lunches. They
would like to keep their costs down, but they also need to meet the
nutritional requirements of the children. They have already decided
to go with peanut butter and jelly sandwiches, and some combination of graham crackers, milk, and orange juice. The nutritional
content of each food choice and its cost are given in the table below.
Food Item
Bread (1 slice)
Peanut butter (1 tbsp.)
Strawberry jelly (1 tbsp.)
Graham cracker (1 cracker)
Milk (1 cup)
Juice (1 cup)
a. Identify the requirements that lead to resource constraints, to benefit constraints, and to fixed-requirement constraints.
b. Formulate and solve a linear programming model
for this problem on a spreadsheet.
c. Express the model in algebraic form.
3.27.
The Cost-Less Corp. supplies its four retail outlets from
its four plants. The shipping cost per shipment from each plant to
each retail outlet is given below.
Unit Shipping Cost
Retail Outlet:
Plant
1
2
3
4
1
2
3
4
$500
200
300
200
$600
900
400
100
$400
100
200
300
$200
300
100
200
Plants 1, 2, 3, and 4 make 10, 20, 20, and 10 shipments per
month, respectively. Retail outlets 1, 2, 3, and 4 need to receive
20, 10, 10, and 20 shipments per month, respectively.
Calories
from Fat
Total
Calories
Vitamin C
(mg)
Protein
(g)
Cost
(¢)
10
75
0
20
70
0
70
100
50
60
150
100
0
0
3
0
2
120
3
4
0
1
8
1
5
4
7
8
15
35
The nutritional requirements are as follows. Each child
should receive between 400 and 600 calories. No more than 30
percent of the total calories should come from fat. Each child
should consume at least 60 milligrams (mg) of vitamin C and
The distribution manager, Randy Smith, now wants to determine the best plan for how many shipments to send from each
plant to the respective retail outlets each month. Randy’s objective is to minimize the total shipping cost.
Chapter 3 Problems 113
Formulate this problem as a transportation problem on a
spreadsheet and then use Solver to obtain an optimal solution.
3.28.
The Childfair Company has three plants producing
child push chairs that are to be shipped to four distribution centers. Plants 1, 2, and 3 produce 12, 17, and 11 shipments per
month, respectively. Each distribution center needs to receive
10 shipments per month. The distance from each plant to the
respective distribution centers is given below.
Distance to Distribution Center (Miles)
Plant
1
2
3
1
2
3
4
800
1,100
600
1,300
1,400
1,200
400
600
800
700
1,000
900
The freight cost for each shipment is $100 plus 50 cents/mile.
How much should be shipped from each plant to each of the
distribution centers to minimize the total shipping cost?
Formulate this problem as a transportation problem on a
spreadsheet and then use Solver to obtain an optimal solution.
3.29.
The Onenote Co. produces a single product at three
plants for four customers. The three plants will produce 60, 80,
and 40 units, respectively, during the next week. The firm has
made a commitment to sell 40 units to customer 1, 60 units to
customer 2, and at least 20 units to customer 3. Both customers
3 and 4 also want to buy as many of the remaining units as possible. The net profit associated with shipping a unit from plant i
for sale to customer j is given by the following table.
Customer
Plant
1
2
3
1
2
3
4
$800
500
600
$700
200
400
$500
100
300
$200
300
500
Management wishes to know how many units to sell to customers 3 and 4 and how many units to ship from each of the
plants to each of the customers to maximize profit. Formulate
and solve a spreadsheet model for this problem.
3.30.
The Move-It Company has two plants building forklift
trucks that then are shipped to three distribution centers. The
production costs are the same at the two plants, and the cost of
shipping each truck is shown below for each combination of
plant and distribution center.
Distribution Center
Plant
A
B
1
2
3
$800
600
$700
800
$400
500
A total of 60 forklift trucks are produced and shipped per
week. Each plant can produce and ship any amount up to a maximum of 50 trucks per week, so there is considerable flexibility
on how to divide the total production between the two plants so
as to reduce shipping costs. However, each distribution center
must receive exactly 20 trucks per week.
Management’s objective is to determine how many forklift
trucks should be produced at each plant, and then what the overall shipping pattern should be to minimize total shipping cost.
Formulate and solve a spreadsheet model for this problem.
3.31.
Redo Problem 3.30 when any distribution center may
receive any quantity between 10 and 30 forklift trucks per week
in order to further reduce total shipping cost, provided only that
the total shipped to all three distribution centers must still equal
60 trucks per week.
3.32.
Read the referenced article that fully describes the management science study summarized in the application vignette
presented in Section 3.6. Briefly describe how the model for the
assignment problem was applied in this study. Then list the various
financial and nonfinancial benefits that resulted from this study.
3.33.
Consider the assignment problem having the following
cost table.
Job
Person
A
B
C
1
2
3
$5
3
2
$7
6
3
$4
5
4
The optimal solution is A-3, B-1, C-2, with a total cost of $10.
Formulate this problem on a spreadsheet and then use Solver
to obtain the optimal solution identified above.
3.34.
Four cargo ships will be used for shipping goods from
one port to four other ports (labeled 1, 2, 3, 4). Any ship can be
used for making any one of these four trips. However, because
of differences in the ships and cargoes, the total cost of loading,
transporting, and unloading the goods for the different ship–port
combinations varies considerably, as shown in the following table.
Port
Ship
1
2
3
4
1
2
3
4
$500
600
700
500
$400
600
500
400
$600
700
700
600
$700
500
600
600
The objective is to assign the four ships to four different ports
in such a way as to minimize the total cost for all four shipments.
a. Describe how this problem fits into the format for
an assignment problem.
b. Formulate and solve this problem on a spreadsheet.
3.35.
Reconsider Problem 3.30. Now distribution centers
1, 2, and 3 must receive exactly 10, 20, and 30 units per week,
114 Chapter Three Linear Programming: Formulation and Applications
respectively. For administrative convenience, management has
decided that each distribution center will be supplied totally by a
single plant, so that one plant will supply one distribution center
and the other plant will supply the other two distribution centers.
The choice of these assignments of plants to distribution centers is
to be made solely on the basis of minimizing total shipping cost.
Formulate and solve a spreadsheet model for this problem.
3.36.
Vincent Cardoza is the owner and manager of a machine
shop that does custom order work. This Wednesday afternoon,
he has received calls from two customers who would like to
place rush orders. One is a trailer hitch company that would like
some custom-made heavy-duty tow bars. The other is a minicar-carrier company that needs some customized stabilizer bars.
Both customers would like as many as possible by the end of
the week (two working days). Since both products would require
the use of the same two machines, Vincent needs to decide and
inform the customers this afternoon about how many of each
product he will agree to make over the next two days.
Each tow bar requires 3.2 hours on machine 1 and 2 hours
on machine 2. Each stabilizer bar requires 2.4 hours on machine
1 and 3 hours on machine 2. Machine 1 will be available for
16 hours over the next two days and machine 2 will be available
for 15 hours. The profit for each tow bar produced would be $130
and the profit for each stabilizer bar produced would be $150.
Vincent now wants to determine the mix of these production
quantities that will maximize the total profit.
a. Formulate an integer programming model in algebraic form for this problem.
b. Formulate and solve the model on a spreadsheet.
3.37.
Pawtucket University is planning to buy new copier
machines for its library. Three members of its Management
Science Department are analyzing what to buy. They are considering two different models: Model A, a high-speed copier,
and Model B, a lower speed but less expensive copier. Model A
can handle 20,000 copies a day and costs $6,000. Model B can
handle 10,000 copies a day but only costs $4,000. They would
like to have at least six copiers so that they can spread them
throughout the library. They also would like to have at least one
high-speed copier. Finally, the copiers need to be able to handle
a capacity of at least 75,000 copies per day. The objective is to
determine the mix of these two copiers that will handle all these
requirements at minimum cost.
a. Formulate and solve a spreadsheet model for this
problem.
b. Formulate this same model in algebraic form.
3.38.
Northeastern Airlines is considering the purchase of
new long-, medium-, and short-range jet passenger airplanes.
The purchase price would be $201 million for each longrange plane, $150 million for each medium-range plane, and
$105 million for each short-range plane. The board of directors has authorized a maximum commitment of $4.5 billion for
these purchases. Regardless of which airplanes are purchased,
air travel of all distances is expected to be sufficiently large that
these planes would be utilized at essentially maximum capacity. It is estimated that the net annual profit (after capital recovery costs are subtracted) would be $12.6 million per long-range
plane, $9 million per medium-range plane, and $6.9 million per
short-range plane.
It is predicted that enough trained pilots will be available
to the company to crew 30 new airplanes. If only short-range
planes were purchased, the maintenance facilities would be able
to handle 40 new planes. However, each medium-range plane is
11
equivalent to ___
​​   ​​short-range planes, and each long-range plane
3
12
is equivalent to ___
​​   ​​short-range planes in terms of their use of the
3
maintenance facilities.
The information given here was obtained by a preliminary
analysis of the problem. A more detailed analysis will be conducted subsequently. However, using the preceding data as a first
approximation, management wishes to know how many planes
of each type should be purchased to maximize profit.
a. Formulate and solve a spreadsheet model for this
problem.
b. Formulate this model in algebraic form.
Case 3-1
Shipping Wood to Market
Alabama Atlantic is a lumber company that has three sources of
wood and five markets to be supplied. The annual availability of
wood at sources 1, 2, and 3 is 15, 20, and 15 million board feet,
respectively. The amount that can be sold annually at markets 1, 2,
3, 4, and 5 is 11, 12, 9, 10, and 8 million board feet, respectively.
Unit Cost by Rail ($1,000s) to Market
Source
1
2
3
In the past, the company has shipped the wood by train.
However, because shipping costs have been increasing, the
alternative of using ships to make some of the deliveries is
being investigated. This alternative would require the company
to invest in some ships. Except for these investment costs, the
Unit Cost by Ship ($1,000s) to Market
1
2
3
4
5
1
2
3
4
5
61
69
59
72
78
66
45
60
63
55
49
61
66
56
47
31
36
—
38
43
33
24
28
36
—
24
32
35
31
26
Case 3-2 Capacity Concerns 115
shipping costs in thousands of dollars per million board feet by
rail and by water (when feasible) is given by the above table for
each route.
The capital investment (in thousands of dollars) in ships
required for each million board feet to be transported annually by
ship along each route is given next.
Unit Investment for Ships ($1,000s) to
Market
Source
1
2
3
4
5
1
2
3
275
293
—
303
318
283
238
270
275
—
250
268
285
265
240
Considering the expected useful life of the ships and the time
value of money, the equivalent uniform annual cost of these investments is one-tenth the amount given in the table. The objective
is to determine the overall shipping plan that minimizes the total
equivalent uniform annual cost (including shipping costs).
You are the head of the management science team that has
been assigned the task of determining this shipping plan for each
of the three options listed next.
Option 1: Continue shipping exclusively by rail.
Option 2: Switch to shipping exclusively by water (except
where only rail is feasible).
Option 3: Ship by either rail or water, depending on which is
less expensive for the particular route.
Present your results for each option. Compare.
Finally, consider the fact that these results are based on current shipping and investment costs, so that the decision on the
option to adopt now should take into account management’s
projection of how these costs are likely to change in the future.
For each option, describe a scenario of future cost changes that
would justify adopting that option now.
Case 3-2
Capacity Concerns
Bentley Hamilton throws the business section of The New York
Times onto the conference room table and watches as his associates jolt upright in their overstuffed chairs.
Mr. Hamilton wants to make a point.
He throws the front page of the The Wall Street Journal on
top of The New York Times and watches as his associates widen
their eyes once heavy with boredom.
Mr. Hamilton wants to make a big point.
He then throws the front page of the Financial Times on top
of the newspaper pile and watches as his associates dab the fine
beads of sweat off their brows.
Mr. Hamilton wants his point indelibly etched into his associates’ minds.
“I have just presented you with three leading financial
newspapers carrying today’s top business story,” Mr. H
­ amilton
declares in a tight, angry voice. “My dear associates, our company is going to hell in a hand basket! Shall I read you the headlines? From The New York Times, ‘CommuniCorp stock drops
to lowest in 52 weeks.’ From The Wall Street Journal, ‘CommuniCorp loses 25 percent of the wireless router market in only
one year.’ Oh, and my favorite, from the Financial Times, ‘CommuniCorp cannot CommuniCate: CommuniCorp stock drops
because of internal communications disarray.’ How did our
company fall into such dire straits?”
Mr. Hamilton next points at a line sloping slightly upward on
the conference room display. “This is a graph of our productivity
over the last 12 months. As you can see from the graph, productivity in our router production facility has increased steadily over the
last year. Clearly, productivity is not the cause of our problem.”
Mr. Hamilton next displays a second graph showing a line
sloping steeply upward. “This is a graph of our missed or late
orders over the last 12 months.” Mr. Hamilton hears an audible
gasp from his associates. “As you can see from the graph, our
missed or late orders have increased steadily and significantly
over the past 12 months. I think this trend explains why we have
been losing market share, causing our stock to drop to its lowest level in 52 weeks. We have angered and lost the business of
retailers, our customers who depend upon on-time deliveries to
meet the demand of consumers.”
“Why have we missed our delivery dates when our productivity level should have allowed us to fill all orders?” Mr. ­Hamilton
asks. “I called several departments to ask this question.”
“It turns out that we have been producing routers for the
hell of it!” Mr. Hamilton says in disbelief. “The marketing and
sales departments do not communicate with the manufacturing department, so manufacturing executives do not know what
routers to produce to fill orders. The manufacturing executives
want to keep the plant running, so they produce routers regardless of whether the routers have been ordered. Finished routers
are sent to the warehouse, but marketing and sales executives
do not know the number and styles of routers in the warehouse.
They try to communicate with warehouse executives to determine if the routers in inventory can fill the orders, but they rarely
receive answers to their questions.”
Mr. Hamilton pauses and looks directly at his associates.
“Ladies and gentlemen, it seems to me that we have a serious
internal communications problem. I intend to correct this problem immediately. I want to begin by installing a companywide
computer network to ensure that all departments have access to
critical documents and are able to communicate with each other
more easily. Because this intranet will represent a large change
from the current communications infrastructure, I expect some
bugs in the system and some resistance from employees. I therefore want to phase in the installation of the intranet.”
Mr. Hamilton passes the following time line and requirements chart to his associates (IN = intranet).
116 Chapter Three Linear Programming: Formulation and Applications
Month 1
Month 2
Month 3
Month 4
Month 5
IN
education
Install IN in
sales
Install IN in
manufacturing
Install IN in
warehouse
Install IN in
marketing
Department
Number of Employees
Sales
Manufacturing
Warehouse
Marketing
60
200
30
75
Mr. Hamilton proceeds to explain the time line and requirements chart. “In the first month, I do not want to bring any
department onto the intranet; I simply want to disseminate information about it and get buy-in from employees. In the second
month, I want to bring the sales department onto the intranet
since the sales department receives all critical information from
customers. In the third month, I want to bring the manufacturing department onto the intranet. In the fourth month, I want
to install the intranet at the warehouse, and in the fifth and
final month, I want to bring the marketing department onto
the intranet. The requirements chart above lists the number of
employees requiring access to the intranet in each department.”
Mr. Hamilton turns to Emily Jones, the head of Corporate
Information Management. “I need your help in planning for the
installation of the intranet. Specifically, the company needs to
purchase servers for the internal network. Employees will connect to company servers and download information to their own
desktop computers.”
Mr. Hamilton passes Emily the following chart detailing the
types of servers available, the number of employees each server
supports, and the cost of each server.
Type of Server
Number of Employees
Server Supports
Mini Desktop Server
Desktop Server
Workstation Server
Full Rack Server
Up to 30 employees
Up to 80 employees
Up to 200 employees
Up to 2,000 employees
“Emily, I need you to decide what servers to purchase and
when to purchase them to minimize cost and to ensure that the
company possesses enough server capacity to follow the intranet
implementation timeline,” Mr. Hamilton says. “For example, you
may decide to buy one large server during the first month to support all employees, or buy several small servers during the first
month to support all employees, or buy one small server each
month to support each new group of employees gaining access
to the intranet.”
“There are several factors that complicate your decision,”
Mr. Hamilton continues. We can get a discount of 10 percent off
each workstation server purchased, but only if we purchase servers in the first or second month. We can get a 25 percent discount
off all full rack servers purchased in the first two months. You are
also limited in the amount of money you can spend during the first
month. CommuniCorp has already allocated much of the budget
for the next two months, so you only have a total of $9,500 available to purchase servers in months 1 and 2. Finally, the manufacturing department requires at least one of the three more powerful
servers. Have your decision on my desk at the end of the week.”
a. Emily first decides to evaluate the number and type of servers
to purchase on a month-to-month basis. For each month, formulate a spreadsheet model to determine which servers Emily
should purchase in that month to minimize costs in that month
and support the new users given your results for the preceding
months. How many and which types of servers should she purchase in each month? How much is the total cost of the plan?
b. Emily realizes that she could perhaps achieve savings if she bought
a larger server in the initial months to support users in the final
months. She therefore decides to evaluate the number and type of
servers to purchase over the entire planning period. Formulate a
spreadsheet model to determine which servers Emily should purchase in which months to minimize total cost and support all new
users. How many and which types of servers should she purchase
in each month? How much is the total cost of the plan?
Cost of Server
$ 2,500
5,000
10,000
25,000
c. Why is the answer using the first method different from that
using the second method?
d. Are there other costs for which Emily is not accounting in her
problem formulation? If so, what are they?
e. What further concerns might the various departments of
CommuniCorp have regarding the intranet?
Case 3-3 Fabrics and Fall Fashions 117
Case 3-3
Fabrics and Fall Fashions
From the 10th floor of her office building, Katherine Rally
watches the swarms of New Yorkers fight their way through the
streets infested with yellow cabs and the sidewalks littered with hot
dog stands. On this sweltering July day, she pays particular attention to the fashions worn by the various women and wonders what
they will choose to wear in the fall. Her thoughts are not simply
random musings; they are critical to her work since she owns and
manages TrendLines, an elite women’s clothing company.
Today is an especially important day because she must meet
with Ted Lawson, the production manager, to decide upon next
month’s production plan for the fall line. Specifically, she must
determine the quantity of each clothing item she should produce
given the plant’s production capacity, limited resources, and
demand forecasts. Accurate planning for next month’s production
is critical to fall sales since the items produced next month will
appear in stores during September and women generally buy the
majority of the fall fashions when they first appear in September.
She turns back to her sprawling glass desk and looks at the
numerous papers covering it. Her eyes roam across the clothing
patterns designed almost six months ago, the lists of material
requirements for each pattern, and the lists of demand forecasts
for each pattern determined by customer surveys at fashion
shows. She remembers the hectic and sometimes nightmarish
days of designing the fall line and presenting it at fashion shows
in New York, Milan, and Paris. Ultimately, she paid her team of
six designers a total of $860,000 for their work on her fall line.
With the cost of hiring runway models, hair stylists, and make-up
artists; sewing and fitting clothes; building the set; choreographing and rehearsing the show; and renting the conference hall, each
of the three fashion shows cost her an additional $2,700,000.
She studies the clothing patterns and material requirements.
Her fall line consists of both professional and casual fashions. She
determined the price for each clothing item by taking into account
the quality and cost of material, the cost of labor and machining, the
demand for the item, and the prestige of the TrendLines brand name.
The fall professional fashions include.
Clothing
Item
Material
Requirements
Price
Labor and
Machine Cost
Tailored
wool slacks
3 yards of wool
2 yards of acetate
for lining
$300
$160
Cashmere
sweater
1.5 yards of
cashmere
450
150
Silk blouse
Silk camisole
1.5 yards of silk
180
100
0.5 yard of silk
120
Tailored skirt
60
2 yards of rayon
270
120
320
140
1.5 yards of acetate for lining
Wool blazer
2.5 yards of wool
1.5 yards of acetate for lining
The fall casual fashions include.
Labor and
Machine
Cost
Clothing
Item
Material
Requirements
Price
Velvet pants
3 yards of velvet
$350
$175
2 yards of acetate
for lining
Cotton
sweater
1.5 yards of cotton
130
60
Cotton
miniskirt
0.5 yard of cotton
75
40
Velvet shirt
1.5 yards of velvet
200
160
Button-down
blouse
1.5 yards of rayon
120
90
She knows that for the next month, she has ordered 45,000
yards of wool, 28,000 yards of acetate, 9,000 yards of cashmere,
18,000 yards of silk, 30,000 yards of rayon, 20,000 yards of velvet, and 30,000 yards of cotton for production. The prices of the
materials are listed below.
Material
Wool
Acetate
Cashmere
Silk
Rayon
Velvet
Cotton
Price per Yard
$ 9.00
1.50
60.00
13.00
2.25
12.00
2.50
Any material that is not used in production can be sent back
to the textile wholesaler for a full refund, although scrap material cannot be sent back to the wholesaler.
She knows that the production of both the silk blouse and cotton sweater leaves leftover scraps of material. Specifically, for
the production of one silk blouse or one cotton sweater, 2 yards
of silk and cotton, respectively, are needed. From these 2 yards,
1.5 yards are used for the silk blouse or the cotton sweater and
0.5 yard is left as scrap material. She does not want to waste the
material, so she plans to use the rectangular scrap of silk or cotton to produce a silk camisole or cotton miniskirt, respectively.
Therefore, whenever a silk blouse is produced, a silk camisole is
also produced. Likewise, whenever a cotton sweater is produced,
a cotton miniskirt is also produced. Note that it is possible to
produce a silk camisole without producing a silk blouse and a
cotton miniskirt without producing a cotton sweater.
The demand forecasts indicate that some items have limited demand. Specifically, because the velvet pants and velvet
shirts are fashion fads, TrendLines has forecasted that it can
sell only 5,500 pairs of velvet pants and 6,000 velvet shirts.
TrendLines does not want to produce more than the forecasted
demand because once the pants and shirts go out of style, the
118 Chapter Three Linear Programming: Formulation and Applications
company cannot sell them. TrendLines can produce less than the
forecasted demand, however, since the company is not required
to meet the demand. The cashmere sweater also has limited
demand because it is quite expensive, and TrendLines knows it
can sell at most 4,000 cashmere sweaters. The silk blouses and
camisoles have limited demand because many women think silk
is too hard to care for, and TrendLines projects that it can sell at
most 12,000 silk blouses and 15,000 silk camisoles.
The demand forecasts also indicate that the wool slacks,
tailored skirts, and wool blazers have a great demand because
they are basic items needed in every professional wardrobe.
Specifically, the demand is 7,000 pairs of wool slacks and 5,000
wool blazers. Katherine wants to meet at least 60 percent of the
demand for these two items to maintain her loyal customer base
and not lose business in the future. Although the demand for
tailored skirts could not be estimated, Katherine feels she should
make at least 2,800 of them.
a. Ted is trying to convince Katherine not to produce any velvet
shirts since the demand for this fashion fad is quite low. He
argues that this fashion fad alone accounts for $500,000 of
the fixed design and other costs. The net contribution (price
of clothing item – materials cost – labor cost) from selling
the fashion fad should cover these fixed costs. Each velvet
shirt generates a net contribution of $22. He argues that given
the net contribution, even satisfying the maximum demand
will not yield a profit. What do you think of Ted’s argument?
b. Formulate and solve a linear programming problem to maximize
profit given the production, resource, and demand constraints.
Before she makes her final decision, Katherine plans to
explore the following questions independently, except where
otherwise indicated.
c. The textile wholesaler informs Katherine that the velvet cannot be sent back because the demand forecasts show that the
demand for velvet will decrease in the future. Katherine can
therefore get no refund for the velvet. How does this fact
change the production plan?
d. What is an intuitive economic explanation for the difference
between the solutions found in parts b and c?
e. The sewing staff encounters difficulties sewing the arms and
lining into the wool blazer since the blazer pattern has an
awkward shape and the heavy wool material is difficult to
cut and sew. The increased labor time to sew a wool blazer
increases the labor and machine cost for each blazer by $80.
Given this new cost, how many of each clothing item should
TrendLines produce to maximize profit?
f. The textile wholesaler informs Katherine that since another
textile customer canceled his order, she can obtain an extra
10,000 yards of acetate. How many of each clothing item
should TrendLines now produce to maximize profit?
g. TrendLines assumes that it can sell every item that was not
sold during September and October in a big sale in November
at 60 percent of the original price. Therefore, it can sell all
items in unlimited quantity during the November sale. (The
previously mentioned upper limits on demand only concern
the sales during September and October.) What should the
new production plan be to maximize profit?
Case 3-4
New Frontiers
Rob Richman, president of AmeriBank, takes off his glasses,
rubs his eyes in exhaustion, and squints at the clock in his study.
It reads 3 AM. For the last several hours, Rob has been poring over
­AmeriBank’s financial statements from the last three quarters
of operation. AmeriBank, a medium-sized bank with branches
throughout the United States, is headed for dire economic straits.
The bank, which provides transaction, savings, investment, and
loan services, has been experiencing a steady decline in its net
income over the past year, and trends show that the decline will
continue. The bank is simply losing customers to nonbank and
foreign bank competitors.
AmeriBank is not alone in its struggle to stay out of the
red. From his daily industry readings, Rob knows that many
­American banks have been suffering significant losses because
of increasing competition from nonbank and foreign bank competitors offering services typically in the domain of American
banks. Because the nonbank and foreign bank competitors specialize in particular services, they are able to better capture the
market for those services by offering less expensive, more efficient, more convenient services. For example, large corporations
now turn to foreign banks and commercial paper offerings for
loans, and affluent Americans now turn to money-market funds
for investment. Banks face the daunting challenge of distinguishing themselves from nonbank and foreign bank competitors.
Rob has concluded that one strategy for distinguishing
AmeriBank from its competitors is to improve services that
­
nonbank and foreign bank competitors do not readily provide:
transaction services. He has decided that a more convenient transaction method must logically succeed the automatic teller machine,
and he believes that electronic banking over the Internet allows
this convenient transaction method. Over the Internet, customers
are able to perform transactions on their desktop computers either
at home or work. The explosion of the Internet means that most
potential customers understand and use it. He therefore feels that if
AmeriBank offers Web banking (as the practice of Internet banking is commonly called), the bank will attract many new customers.
Before Rob undertakes the project to make Web banking
possible, however, he needs to understand the market for Web
banking and the services AmeriBank should provide over the
Internet. For example, should the bank only allow customers to
access account balances and historical transaction information
over the Internet, or should the bank develop a strategy to allow
customers to make deposits and withdrawals over the Internet?
Should the bank try to recapture a portion of the investment market by continuously running stock prices and allowing customers
to make stock transactions over the Internet for a minimal fee?
Therefore, Rob has concluded that a major survey project should
be undertaken to learn what customers want.
Because AmeriBank is not in the business of performing
surveys, Rob has decided to outsource the survey project to a
professional survey company. He has opened the project up for
bidding by several survey companies and will award the project
Case 3-5 Assigning Students to Schools 119
to the company that is willing to perform the survey for the least
cost. Rob provided each survey company with a list of survey
requirements to ensure that AmeriBank receives the needed
information for planning the Web banking project.
Because different age groups require different services, AmeriBank is interested in surveying four different age groups. The first
group encompasses customers who are 18 to 25 years old. The
bank assumes that this age group has limited yearly income and
performs minimal transactions. The second group encompasses
customers who are 26 to 40 years old. This age group has significant sources of income, performs many transactions, requires
numerous loans for new houses and cars, and invests in various
securities. The third group encompasses customers who are 41
to 50 years old. These customers typically have the same level of
income and perform the same number of transactions as the s­ econd
age group, but the bank assumes that these customers are less
likely to use Web banking since they have not become as comfortable with the explosion of computers or the Internet. Finally, the
fourth group encompasses customers who are 51 years of age and
over. These customers commonly crave security and require continuous information on retirement funds. The bank believes that it
is highly unlikely that many customers in this age group will use
Web banking, but the bank desires to learn the needs of this age
group for the future. AmeriBank wants to interview 2,000 customers with at least 20 percent from the first age group, at least
27.5 percent from the second age group, at least 15 percent from the
third age group, and at least 15 percent from the fourth age group.
Rob understands that some customers are uncomfortable
with using the Internet. He therefore wants to ensure that the
survey includes a mix of customers who know the Internet well
and those that have less exposure to the Internet. To ensure
that AmeriBank obtains the correct mix, he wants to interview
at least 15 percent of customers from the Silicon Valley where
Internet use is high, at least 35 percent of customers from big
cities where Internet use is medium, and at least 20 percent of
customers from small towns where Internet use is low.
Sophisticated Surveys is one of three survey companies
competing for the project. It has performed an initial analysis
of these survey requirements to determine the cost of surveying
different populations. The costs per person surveyed are listed in
the following table.
Age Group
Region
Silicon Valley
Big cities
Small towns
18 to 25 26 to 40 41 to 50 51 and over
$4.75
5.25
6.50
$6.50
5.75
7.50
$6.50
6.25
7.50
$5.00
6.25
7.25
Sophisticated Surveys explores the following options
cumulatively.
a. Formulate a linear programming model to minimize costs
while meeting all survey constraints imposed by AmeriBank.
b. If the profit margin for Sophisticated Surveys is 15 percent of
cost, what bid will it submit?
c. After submitting its bid, Sophisticated Surveys is informed
that it has the lowest cost but that AmeriBank does not like
the solution. Specifically, Rob feels that the selected survey
population is not representative enough of the banking customer population. Rob wants at least 50 people of each age
group surveyed in each region. What is the new bid made by
Sophisticated Surveys?
d. Rob feels that Sophisticated Surveys oversampled the 18-to25-year-old population and the Silicon Valley population. He
imposes a new constraint that no more than 600 individuals
can be surveyed from the 18-to-25-year-old population and
no more than 650 individuals can be surveyed from the Silicon Valley population. What is the new bid?
e. When Sophisticated Surveys calculated the cost of reaching
and surveying particular individuals, the company thought
that reaching individuals in young populations would be
easiest. In a recently completed survey, however, Sophisticated Surveys learned that this assumption was wrong. The
new costs for surveying the 18-to-25-year-old population are
listed below.
Region
Silicon Valley
Big cities
Small towns
Cost per Person
$6.50
6.75
7.00
Given the new costs, what is the new bid?
f. To ensure the desired sampling of individuals, Rob imposes
even stricter requirements. He fixes the exact percentage of
people that should be surveyed from each population. The
requirements are listed next.
Population
18 to 25
26 to 40
41 to 50
51 and over
Silicon Valley
Big cities
Small towns
Percentage of People Surveyed
25%
35
20
20
20
50
30
By how much would these new requirements increase the cost
of surveying for Sophisticated Surveys? Given the 15 ­percent
profit margin, what would Sophisticated Surveys bid?
Case 3-5
Assigning Students to Schools
The Springfield School Board has made the decision to close
one of its middle schools (sixth, seventh, and eighth grades) at
the end of this school year and reassign all of next year’s middle
school students to the three remaining middle schools. The
school district provides busing for all middle school students
who must travel more than approximately a mile, so the school
120 Chapter Three Linear Programming: Formulation and Applications
board wants a plan for reassigning the students that will minimize the total busing cost. The annual cost per student for busing
from each of the six residential areas of the city to each of the
schools is shown in the following table (along with other basic
data for next year), where 0 indicates that busing is not needed
and a dash indicates an infeasible assignment.
How much does this increase the total busing cost? (This line
of analysis will be pursued more rigorously in Case 7-3.)
The school board is considering eliminating some busing to
reduce costs. Option 1 is to only eliminate busing for students traveling 1 to 1.5 miles, where the cost per student is given in the table
Busing Cost per Student
Area
Number of
Students
Percentage
in 6th Grade
Percentage
in 7th Grade
Percentage
in 8th Grade
School 1
School 2
School 3
450
600
550
350
500
450
32
37
30
28
39
34
38
28
32
40
34
28
30
35
38
32
27
38
$300
—
600
200
0
500
$ 0
400
300
500
—
300
$700
500
200
—
400
0
School capacity:
900
1,100
1,000
1
2
3
4
5
6
The school board also has imposed the restriction that
each grade must constitute between 30 and 36 percent of each
school’s population. The above table shows the percentage of
each area’s middle school population for next year that falls into
each of the three grades. The school attendance zone boundaries can be drawn so as to split any given area among more than
one school, but assume that the percentages shown in the table
will continue to hold for any partial assignment of an area to a
school.
You have been hired as a management science consultant
to assist the school board in determining how many students in
each area should be assigned to each school.
a. Formulate and solve a linear programming model for this
problem.
b. What is your resulting recommendation to the school board?
After seeing your recommendation, the school board
expresses concern about all the splitting of residential areas
among multiple schools. They indicate that they “would like to
keep each neighborhood together.”
c. Adjust your recommendation as well as you can to enable
each area to be assigned to just one school. (Adding this
restriction may force you to fudge on some other constraints.)
as $200. Option 2 is to also eliminate busing for students traveling
1.5 to 2 miles, where the estimated cost per student is $300.
d. Revise the model from part a to fit Option 1, and solve. Compare these results with those from part b, including the reduction in total busing cost.
e. Repeat part d for Option 2.
The school board now needs to choose among the three alternative busing plans (the current one or Option 1 or Option 2).
One important factor is busing costs. However, the school board
also wants to place equal weight on a second factor: the inconvenience and safety problems caused by forcing students to travel
by foot or bicycle a substantial distance (more than a mile, and
especially more than 1.5 miles). Therefore, they want to choose
a plan that provides the best trade-off between these two factors.
f. Use your results from parts b, d, and e to summarize the key
information related to these two factors that the school board
needs to make this decision.
g. Which decision do you think should be made? Why?
Note: This case will be continued in later chapters (Cases 5-4
and 7-3), so we suggest that you save your analysis, including
your basic spreadsheet model.
Case 3-6
Reclaiming Solid Wastes
The Save-It Company operates a reclamation center that collects four types of solid waste materials and then treats them
so that they can be amalgamated (treating and amalgamating
are separate processes) into a salable product. Three different grades of this product can be made, depending on the mix
of the materials used. (See the first table.) Although there is
some flexibility in the mix for each grade, quality standards
specify the minimum or maximum amount of the materials
allowed in that product grade. (This minimum or maximum
amount is the weight of the material expressed as a percentage
of the total weight for that product grade.) For each of the
two higher grades, a fixed percentage is specified for one of
the materials. These specifications are given in the first table
along with the cost of amalgamation and the selling price for
each grade.
The reclamation center collects its solid waste materials
from some regular sources and so is normally able to maintain a
steady rate for treating them. The second table gives the quantities available for collection and treatment each week, as well as
the cost of treatment, for each type of material.
Case 3-7 Project Pickings 121
Grade
A
Specification
Material 1: Not more than 30% of the total
Material 2: Not less than 40% of the total
Material 3: Not more than 50% of the total
Material 4: Exactly 20% of the total
Amalgamation
Cost per Pound
Selling Price
per Pound
$3.00
$8.50
B
Material 1: Not more than 50% of the total
Material 2: Not less than 10% of the total
Material 4: Exactly 10% of the total
2.50
7.00
C
Material 1: Not more than 70% of the total
2.00
5.50
Material
1
2
3
4
Available
Pounds/Week
Treatment Cost
per Pound
3,000
2,000
4,000
1,000
$3.00
6.00
4.00
5.00
The Save-It Company is solely owned by Green Earth,
an organization that is devoted to dealing with environmental issues; Save-It’s profits are all used to help support Green
Earth’s activities. Green Earth has raised contributions and
grants, amounting to $30,000 per week, to be used exclusively to
cover the entire treatment cost for the solid waste materials. The
board of directors of Green Earth has instructed the management
of Save-It to divide this money among the materials in such a
way that at least half of the amount available of each material is
actually collected and treated. These additional restrictions are
listed in the second table.
Within the restrictions specified in the two tables, management wants to allocate the materials to product grades so as to
Additional Restrictions
1. For each material, at least half
of the pounds/week available
should be collected and
treated.
2. $30,000 per week should be
used to treat these materials.
maximize the total weekly profit (total sales income minus total
amalgamation cost).
a. Formulate this problem in linear programming terms by
identifying all the activities, resources, benefits, and fixed
requirements that lurk within it.
b. Formulate and solve a spreadsheet model for this linear programming problem.
c. Express this linear programming model in the spreadsheet in
algebraic form.
Case 3-7
Project Pickings
Tazer, a pharmaceutical manufacturing company, entered the
pharmaceutical market 15 years ago with the introduction of six
new drugs. Five of the six drugs were simply permutations of
existing drugs and therefore did not sell very heavily. The sixth
drug, however, addressed hypertension and was a huge success.
Since Tazer had a patent on the hypertension drug, it experienced
no competition, and profits from the hypertension drug alone
kept Tazer in business. Pharmaceutical patents remain in force
for 20 years, so this one has five more years before it expires.
During the past 15 years, Tazer continued a moderate amount
of research and development, but it never stumbled upon a drug
as successful as the hypertension drug. One reason is that the
company never had the motivation to invest heavily in innovative
research and development. The company was riding the profit wave
generated by its hypertension drug and did not feel the need to
commit significant resources to finding new drug breakthroughs.
Now Tazer is beginning to fear the pressure of competition.
Tazer knows that once the patent expires in five years, generic
drug manufacturing companies will swarm into the market like
vultures. Historical trends show that generic drugs decrease
sales of branded drugs by 75 percent.
Tazer is therefore looking to invest significant amounts of
money in research and development this year to begin the search
for a new breakthrough drug that will offer the company the
same success as the hypertension drug. Tazer believes that if
the company begins extensive research and development now,
the probability of finding a successful drug shortly after the
expiration of the hypertension patent will be high.
As head of research and development at Tazer, you are
responsible for choosing potential projects and assigning project
directors to lead each of the projects. After researching the needs
of the market, analyzing the shortcomings of current drugs,
122 Chapter Three Linear Programming: Formulation and Applications
and interviewing numerous scientists concerning the promising
areas of medical research, you have decided that your department will pursue five separate projects, which are listed below:
Project Up: Develop a more effective antidepressant that
does not cause serious mood swings.
Project Stable: Develop a drug that addresses
manic-depression.
Project Choice: Develop a less intrusive birth control
method for women.
Project Hope: Develop a vaccine to prevent HIV infection.
Project Release: Develop a more effective drug to lower
blood pressure.
For each of the five projects, you are only able to specify the
medical ailment the research should address since you do not
know what compounds will exist and be effective without
research.
You also have five senior scientists to lead the five projects.
You know that scientists are very temperamental people and will
only work well if they are challenged and motivated by the project. To ensure that the senior scientists are assigned to projects
they find motivating, you have established a bidding system for
the projects. You have given each of the five scientists 1,000 bid
points. They assign bids to each project, giving a higher number
of bid points to projects they most prefer to lead.
The following table provides the bids from the five senior
scientists for the five individual projects.
Project
Project Up
Project Stable
Project Choice
Project Hope
Project Release
Project Up: 20
Project Stable: 450
Project Choice: 451
Project Hope: 39
Project Release: 40
Under these new conditions with just four senior scientists,
which scientists will lead which projects to maximize preferences?
e. Do you support the assignments found in part d? Why or why
not?
f. Now you again consider all five scientists. You decide, however, that several scientists cannot lead certain projects. In particular, Dr. Mickey does not have experience with research on
the immune system, so he cannot lead Project Hope. His family also has a history of manic-depression, and you feel that he
would be too personally involved in Project Stable to serve as
an effective project leader. Dr. Mickey therefore cannot lead
Project Stable. Dr. Kvaal also does not have experience with
research on the immune system and cannot lead Project Hope.
In addition, Dr. Kvaal cannot lead Project Release because he
does not have experience with research on the cardiovascular
system. Finally, Dr. Rollins cannot lead Project Up because
her family has a history of depression and you feel she would
be too personally involved in the project to serve as an effective leader. Because Dr. Mickey and Dr. Kvaal cannot lead
two of the five projects, they each have only 600 bid points.
Dr. Rollins has only 800 bid points because she cannot lead
Dr. Kvaal
Dr. Zuner
Dr. Tsai
Dr. Mickey
Dr. Rollins
100
400
200
200
100
0
200
800
0
0
100
100
100
100
600
267
153
99
451
30
100
33
33
34
800
You decide to evaluate a variety of scenarios you think are
likely.
a. Given the bids, you need to assign one senior scientist to each
of the five projects to maximize the preferences of the scientists. What are the assignments?
b. Dr. Rollins is being courted by Harvard Medical School to
accept a teaching position. You are fighting desperately to
keep her at Tazer, but the prestige of Harvard may lure her
away. If this were to happen, the company would give up the
project with the least enthusiasm. Which project would not
be done?
c. If Dr. Rollins leaves, you do not want to sacrifice any project
since researching only four projects decreases the probability
of finding a breakthrough new drug. You decide that either
Dr. Zuner or Dr. Mickey could lead two projects. Under these
new conditions with just four senior scientists, which scientists will lead which projects to maximize preferences?
d. After Dr. Zuner was informed that she and Dr. Mickey are
being considered for two projects, she decided to change
her bids. Dr. Zuner’s new bids for each of the projects
are shown next.
one of the five projects. The following table provides the new
bids of Dr. Mickey, Dr. Kvaal, and Dr. Rollins.
Project
Project Up
Project Stable
Project Choice
Project Hope
Project Release
Dr. Mickey
Dr. Kvaal
Dr. Rollins
300
Can’t lead
125
Can’t lead
175
86
343
171
Can’t lead
Can’t lead
Can’t lead
50
50
100
600
Which scientists should lead which projects to maximize
preferences?
g. You decide that Project Hope and Project Release are too
complex to be led by only one scientist. Therefore, each of
these projects will be assigned two scientists as project leaders. You decide to hire two more scientists in order to staff
all projects: Dr. Arriaga and Dr. Santos. Because of religious
reasons, neither of them want to lead Project Choice and so
they assign 0 bid points to this project. The next table lists all
projects, scientists, and their bids.
Case 3-7 Additional Cases 123
Project
Dr. Kvaal
Dr. Zuner
Dr. Tsai
Dr. Mickey
Dr. Rollins
Dr. Arriaga
Dr. Santos
Project Up
Project Stable
Project Choice
Project Hope
Project Release
86
343
171
Can’t lead
Can’t lead
0
200
800
0
0
100
100
100
100
600
300
Can’t lead
125
Can’t lead
175
Can’t lead
50
50
100
600
250
250
0
250
250
111
1
0
333
555
Which scientists should lead which projects to maximize
preferences?
h. Do you think it is wise to base your decision in part g
only on an optimal solution for a variant of an assignment
problem?
Additional Cases
Additional cases for this chapter are also available at the University of Western Ontario Ivey School of Business website, cases.
ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book.
Chapter Four
The Art of Modeling
with Spreadsheets
Learning Objectives
After completing this chapter, you should be able to
1. Describe the general process for modeling in spreadsheets.
2. Describe some guidelines for building good spreadsheet models.
3. Apply both the general process for modeling in spreadsheets and the guidelines in this
chapter to develop your own spreadsheet model from a description of the problem.
4. Identify some deficiencies in a poorly formulated spreadsheet model.
5. Apply a variety of techniques for debugging a spreadsheet model.
124
Nearly all managers now make extensive use of spreadsheets to analyze business problems.
What they are doing is modeling with spreadsheets.
Spreadsheet modeling is a major emphasis throughout this book. Section 1.2 in
Chapter 1 introduced a spreadsheet model for performing break-even analysis. Section 2.2
in ­Chapter 2 described how to use spreadsheets to formulate linear programming models.
Chapter 3 focused on spreadsheet models for five key categories of linear programming
problems: resource-allocation problems, cost-benefit-trade-off problems, mixed problems,
transportation problems, and assignment problems. Many kinds of spreadsheet models are
discussed in subsequent chapters as well. However, those presentations focus mostly on the
characteristics of spreadsheet models that fit the management science techniques (such as
linear programming) being covered in those chapters. We devote this chapter instead to the
general process of building models with spreadsheets.
Modeling in spreadsheets is more an art than a science. There is no systematic procedure
that invariably will lead to a single correct spreadsheet model. For example, if two people
were given exactly the same business problem to analyze with a spreadsheet, their spreadsheet
models would likely look quite different. There is no one right way of modeling any given
problem. However, some models will be better than others.
Although no completely systematic procedure is available for modeling in spreadsheets,
there is a general process that should be followed. This process has four major steps: (1) plan
the spreadsheet model, (2) build the model, (3) test the model, and (4) analyze the model and
its results. After introducing a case study in Section 4.1, the next section will describe this
plan-build-test-analyze process in some detail and illustrate the process in the context of the
case study. Section 4.2 also will discuss some ways of overcoming common stumbling blocks
in the modeling process.
Unfortunately, despite its logical approach, there is no guarantee that the plan-build-testanalyze process will lead to a “good” spreadsheet model. A good spreadsheet model is easy
to understand, easy to debug, and easy to modify. Section 4.3 presents some guidelines for
building such models. This section also uses the case study in Section 4.1 to illustrate the difference between appropriate formulations and poor formulations of a model.
Even with an appropriate formulation, the initial versions of large spreadsheet models
commonly will include some small but troublesome errors, such as inaccurate references
to cell addresses or typographical errors when entering equations into cells. Indeed, some
4.1 A Case Study: The Everglade Golden Years Company Cash Flow Problem 125
studies have shown that a surprisingly large number of errors typically occur in the first version of such models. These errors often can be difficult to track down. Section 4.4 presents
some helpful ways to debug a spreadsheet model and root out such errors.
The overriding goal of this chapter is to provide a solid foundation for becoming a successful spreadsheet modeler. However, this chapter by itself will not turn you into a highly skilled
modeler. Ultimately, to reach this point, you also will need to study various examples of good
spreadsheet models in the different areas of management science and then have lots of practice in
formulating your own models. This process will continue throughout the remainder of this book.
4.1 A CASE STUDY: THE EVERGLADE GOLDEN YEARS COMPANY
CASH FLOW PROBLEM
With only $1 million in
cash reserves and negative
cash flows looming soon,
loans will be needed to satisfy the company policy of
maintaining a balance of at
least $500,000 at all times.
The Everglade Golden Years Company operates upscale retirement communities in certain
parts of southern Florida. The company was founded in 1946 by Alfred Lee, who was in the
right place at the right time to enjoy many successful years during the boom in the Florida
economy as many wealthy retirees flooded into the area. Today, the company continues to be
run by the Lee family, with Alfred’s grandson, Sheldon Lee, as the CEO.
The past few years have been difficult ones for Everglade. The demand for retirement community housing has been light and Everglade has been unable to maintain full occupancy.
However, this market has picked up recently and the future is looking brighter. Everglade has
recently broken ground for the construction of a new retirement community and has more new
construction planned over the next 10 years (2018 through 2027).
Julie Lee is the chief financial officer (CFO) at Everglade. She has spent the last week in
front of her computer trying to come to grips with the company’s imminent cash flow problem.
Julie has projected Everglade’s net cash flows over the next 10 years as shown in Table 4.1. With
less money currently coming in than would be provided by full occupancy and with all the construction costs for the new retirement community, Everglade will have negative cash flow for the
next few years. With only $1 million in cash reserves, it appears that Everglade will need to take
out some loans in order to meet its financial obligations. Also, to protect against uncertainty,
company policy dictates maintaining a balance of at least $500,000 in cash reserves at all times.
The company’s bank has offered two types of loans to Everglade. The first is a 10-year
loan with interest-only payments made annually and then the entire principal repaid in a single balloon payment after 10 years. The interest rate on this long-term loan is a favorable
5 percent per year. The second option is a series of one-year loans. These loans can be taken
out each year as needed, but each must be repaid (with interest) the following year. Each new
loan can be used to help repay the loan for the preceding year if needed. The interest rate for
these short-term loans currently is projected to be 7 percent per year.
Armed with her cash flow projections and the loan options from the bank, Julie schedules
a meeting with the CEO, Sheldon Lee. Their discussion is as follows:
Julie: Well, we really seem to be in a pickle. There is no way to meet our cash flow problems
without borrowing money.
TABLE 4.1
Projected Net Cash
Flows for the Everglade
Golden Years Company
over the Next Ten Years
Year
Projected Net Cash Flow
(millions of dollars)
2018
2019
2020
2021
2022
2023
2024
2025
2026
2027
−8
−2
−4
3
6
3
−4
7
−2
10
126 Chapter Four The Art of Modeling with Spreadsheets
The objective is to develop
a financial plan that will
keep the company solvent
and then maximize the cash
balance in 2028, after all
the loans are paid off.
Sheldon: I was afraid of that. What are our options?
Julie: I’ve talked to the bank, and we can take out a 10-year loan with an interest rate of
5 percent, or a series of one-year loans at a projected rate of 7 percent.
Sheldon: Wow. That 5 percent rate sounds good. Can we just borrow all that we need
using the 10-year loan?
Julie: That was my initial reaction as well. However, after looking at the cash flow
­projections, I’m not sure the answer is so clear-cut. While we have negative cash flow for the
next few years, the situation looks much brighter down the road. With a 10-year loan, we are
obligated to keep the loan and make the interest payments for 10 years. The one-year loans
are more flexible. We can borrow the money only in the years we need it. This way we can
save on interest payments in the future.
Sheldon: Okay. I can see how the flexibility of the one-year loans could save us some
money. Those loans also will look better if interest rates come down in future years.
Julie: Or they could go higher instead. There’s no way to predict future interest rates, so
we might as well just plan on the basis of the current projection of 7 percent per year.
Sheldon: Yes, you’re right. So which do you recommend, a 10-year loan or a series of
one-year loans?
Julie: Well, there’s actually another option as well. We could consider a combination of
the two types of loans. We could borrow some money long-term to get the lower interest rate
and borrow some money short-term to retain flexibility.
Sheldon: That sounds complicated. What we want is a plan that will keep us solvent
throughout the 10 years and then leave us with as large a cash balance as possible at the
end of the 10 years after paying off all the loans. I also would like to see the amounts of the
interest payments and when loan payments are due. Could you set this up on a spreadsheet to
figure out the best plan?
Julie: You bet. I’ll try that and get back to you.
Sheldon: Great. Let’s plan to meet again next week when you have your report ready.
You’ll see in the next two sections how Julie carefully develops her spreadsheet model for
this cash flow problem.
Review
Questions
4.2
1. What is the advantage of the long-term loan for Everglade?
2. What is the advantage of the series of short-term loans for Everglade?
3. What is the objective for the financial plan that needs to be developed?
OVERVIEW OF THE PROCESS OF MODELING WITH SPREADSHEETS
Spaghetti code is a term
from computer programming. It refers to computer
code that is not logically
organized and thus jumps
all over the place, so it
is jumbled like a plate of
spaghetti.
You will see later that a linear programming model can be incorporated into a spreadsheet
to solve this problem. However, you also will see that the format of this spreadsheet model
does not fit readily into any of the categories of linear programming models described in
Chapter 3. Even the template given in Figure 3.8 that shows the format for a spreadsheet
model of mixed problems (the broadest category of linear programming problems) does
not help in formulating the model for the current problem. The reason is that the Everglade cash flow management problem is an example of a more complicated type of linear
programming problem (a dynamic problem with many time periods) that requires starting
from scratch in carefully formulating the spreadsheet model. Therefore, this example will
nicely illustrate the process of modeling with spreadsheets when dealing with complicated
problems of any type, including those discussed later in the book that do not fit linear
programming.
When presented with a problem like the Everglade problem, the temptation is to jump right
in, launch Excel, and start entering a model. Resist this urge. Developing a spreadsheet model
without proper planning inevitably leads to a model that is poorly organized and filled with
“spaghetti code.”
4.2 Overview of the Process of Modeling with Spreadsheets 127
FIGURE 4.1
A flow diagram for
the general plan-buildtest-analyze process
for ­modeling with
spreadsheets.
Plan
Build
• Visualize where you want to finish
• Do some calculations by hand
• Sketch out a spreadsheet
Start with a
small-scale model
Expand the model
to full scale
Test
Try different trial solutions to check the logic
Analyze
Evaluate proposed solutions and/or optimize
with Solver
If the solution reveals inadequacies
in the model, return to Plan or Build
A modeler might go back
and forth between the Build
and Test steps several times.
Part of the challenge of planning and developing a spreadsheet model is that there is no
standard procedure to follow. It is more an art than a science. However, to provide you with
some structure as you begin learning this art, we suggest that you follow the modeling process
depicted in Figure 4.1.
As suggested by the figure, the four major steps in this process are to (1) plan, (2) build,
(3) test, and (4) analyze the spreadsheet model. The process mainly flows in this order. However, the two-headed arrows between Build and Test indicate a back-and-forth process where
testing frequently results in returning to the Build step to fix some problems discovered during the Test step. This back and forth movement between Build and Test may occur several
times until the modeler is satisfied with the model. At the same time that this back and forth
movement is occurring, the modeler may be involved with further building of the model. One
strategy is to begin with a small version of the model to establish its basic logic and then, after
testing verifies its accuracy, to expand to a full-scale model. Even after completing the testing
and then the analyzing of the model, the process may return to the Build step or even the Plan
step if the Analysis step reveals inadequacies in the model.
Each of these four major steps may also include some detailed steps. For example,
Figure 4.1 lists three detailed steps within the Plan step. Initially, when dealing with a fairly
complicated problem, it is helpful to take some time to perform each of these detailed steps
manually one at a time. However, as you become more experienced with modeling in spreadsheets, you may find yourself merging some of the detailed steps and quickly performing them
mentally. An experienced modeler often is able to do some of these steps mentally, without
working them out explicitly on paper. However, if you find yourself getting stuck, it is likely
that you are missing a key element from one of the previous detailed steps. You then should go
back a step or two and make sure that you have thoroughly completed those preceding steps.
We now describe the various components of the modeling process in the context of the
Everglade cash flow problem. At the same time, we also point out some common stumbling
blocks encountered while building a spreadsheet model and how these can be overcome.
Plan: Visualize Where You Want to Finish
One common stumbling block in the modeling process occurs right at the very beginning.
Given a complicated situation like the one facing Julie at Everglade, sometimes it can be
difficult to decide how to even get started. At this point, rather than focusing on the model
and how to get it started, it can be helpful to think about what you need when you are done.
For example, in Julie’s case, she would be thinking now about what information she should
128 Chapter Four The Art of Modeling with Spreadsheets
provide in her final report to Sheldon. What kinds of numbers need to be included in the
report? The answer to this question can quickly lead you to the heart of the problem and help
get the modeling process started.
The main question Julie is faced with is which loan, or combination of loans, to use and in
what amounts, so one set of numbers that are clearly needed will be the size of the loans to
take. The long-term loan is taken in a single lump sum. Therefore, a single number is needed
to indicate how much money to borrow now at the long-term rate. The short-term loan can be
taken in any or all of the ten years, so there are ten numbers needed to indicate how much to
borrow at the short-term rate in each given year.
In the report, Julie would also need to assure Sheldon that the business will remain solvent
through the next ten years. Therefore, the cash balances will need to be tracked over the next
ten years to make sure they can maintain the needed minimum balance. The report would
also include information on when loan and interest payments are due and in what amounts.
Finally, any data that have been collected would be included, such as the various interest rates
and projected net cash flows from the business. Table 4.2 summarizes the list of numbers that
Julie would need to include in the final report.
Given this list of numbers that are needed in the end, then you can start to think about how
these numbers would fit into a model to try to find a good answer to the problem. Here are
three questions you should ask yourself:
• What are the decisions that need to be addressed in the model? (The numbers representing
these decisions will need to be displayed as changing cells in the model.)
• What is the objective (the overall measure of performance) that should be pursued when
making these decisions? (The number representing the overall measure of performance
will be the objective cell of the model.)
• What are the other kinds of numbers that need to be included in the spreadsheet model?
(Data will go into data cells. Certain other results will need to be calculated before making
the decisions and these results will go into output cells.)
The decisions that Julie needs to address are the size of the loans to take (the first two sets of
numbers in Table 4.2). These 11 numbers will be the changing cells in the spreadsheet model.
The objective when making these decisions is to maximize the cash balance after ten years
while also keeping the company solvent through ten years. Therefore, the final cash balance
will be the objective cell of the spreadsheet model.
The other numbers in the list represent either data (projected net cash flow from the business each year and loan interest rates) or results (cash balances each year, loan payments each
year, and interest payments each year). The data will go into data cells in the spreadsheet
model, while the results will need to be calculated, and will go into output cells in the spreadsheet model.
It is important to distinguish between the numbers that represent decisions (changing cells)
and those that represent results (output cells). For instance, it may be tempting to include the
cash balances as changing cells. These cells clearly change depending on the decisions made.
However, the cash balances are a result of how much is borrowed, how much is paid, and all
of the other cash flows. They cannot be chosen independently but instead are a function of the
other numbers in the spreadsheet. The distinguishing characteristic of changing cells (the loan
amounts) is that they do not depend on anything else. They represent the independent decisions being made. They impact the other numbers, but not vice versa.
TABLE 4.2
List of numbers that
Julie will need to include
in her final report to
Sheldon
Numbers Needed for Final Report
Size of long-term loan
Size of short-term loans
Cash balance each year
Loan payments each year
Interest payments each year
Projected net cash flows each year
Loan interest rates
4.2 Overview of the Process of Modeling with Spreadsheets 129
At this point, you should
know what changing cells
and output cells are needed.
At this stage in the process, you should have a clear idea of what the answer will look like,
including what and how many changing cells are needed, and what kind of results (output
cells) should be obtained.
Plan: Do Some Calculations by Hand
When building a model, another common stumbling block can arise when trying to enter a
formula in one of the output cells. For example, just how does Julie keep track of the cash balances in the Everglade cash flow problem? What formulas need to be entered? There are a lot
of factors that enter into this calculation, so it is easy to get overwhelmed.
If you are getting stuck at this point, it can be a very useful exercise to do some calculations
by hand. Just pick some numbers for the changing cells and determine with a calculator or
pencil and paper what the results should be. For example, pick some loan amounts for Everglade and then calculate the company’s resulting cash balance at the end of the first couple
of years.
To illustrate, suppose that Everglade takes a long-term loan of $6 million and then adds
short-term loans of $2 million in 2018 and $5 million in 2019. How much cash would the
company have left at the end of 2018 and at the end of 2019?
These two quantities can be calculated by hand as follows. In 2018, Everglade has some
initial money in the bank ($1 million), a negative cash flow from its business operations
(− $8 million), and a cash inflow from the long-term and short-term loans ($6 million and
$2 ­million, respectively). Thus, the ending balance for 2018 would be:
Ending balance (2018) = Starting balance
+ Cash flow (2018)
+ LT loan (2018)
+ ST loan (2018)
$1 million
−$8 million
+$6 million
+$2 million
$1 million
The calculations for the year 2019 are a little more complicated. In addition to the starting
balance left over from 2018 ($1 million), negative cash flow from business operations for
2019 (− $2 million), and a new short-term loan for 2019 ($5 million), the company will need
to make interest payments on its 2018 loans as well as pay back the short-term loan from
2018. The ending balance for 2019 is therefore:
Ending balance (2019) = Starting balance (from end of 2018) $1 million
+ Cash flow (2019)
−$2 million
+ ST loan (2019)
+$5 million
−LT interest payment
−(5%)($6 million)
−ST interest payment
−(7%)($2 million)
−ST loan payback (2018)
−$2 million
$1.56 million
Hand calculations can
clarify what formulas are
needed for the output cells.
Doing calculations by hand can help in a couple of ways. First, it can help clarify what
formula should be entered for an output cell. For instance, looking at the by-hand calculations
above, it appears that the formula for the ending balance for a particular year should be
Ending balance = Starting balance + Cash flow + Loans
​​      ​​​
− Interest payments − Loan paybacks
Once a spreadsheet is laid out, this equation will make it straightforward to enter the proper
cell references in the formula for the ending balance for any year in the spreadsheet model.
The second advantage of using hand calculations is that they can help to verify the spreadsheet model after it has been constructed. By plugging in a long-term loan of $6 million,
along with short-term loans of $2 million in 2018 and $5 million in 2019, into a completed
spreadsheet, the ending balances should be the same as calculated above. If they’re not, this
suggests an error in the spreadsheet model (assuming the hand calculations are correct).
130 Chapter Four The Art of Modeling with Spreadsheets
Plan: Sketch Out a Spreadsheet
Plan where the various
blocks of data cells, changing cells, and output cells
should go on the spreadsheet by sketching your
layout ideas on paper.
Any model typically has a large number of different elements that need to be included on the
spreadsheet. For the Everglade problem, these would include some data cells (interest rates,
starting balance, minimum balances, and cash flows), some changing cells (loan amounts),
and a number of output cells (interest payments, loan paybacks, and ending balances). Therefore, a potential stumbling block can arise when trying to organize and lay out the spreadsheet
model. Where should all the pieces fit on the spreadsheet? How do you begin putting together
the spreadsheet?
Before firing up Excel and blindly entering the various elements, it can be helpful to sketch
a layout of the spreadsheet. Is there a logical way to arrange the elements? A little planning at
this stage can go a long way toward building a spreadsheet that is well organized. Don’t bother
with numbers at this point. Simply sketch out blocks on a piece of paper for the various data
cells, changing cells, and output cells, and label them. Concentrate on the layout.
When sketching the layout, a key question for each block of numbers is whether it should
be laid out in a row or a column, or as a two-dimensional table? When addressing this question for the various blocks of numbers, you should also ask whether there are common row
or column headings for different blocks of cells? If so, try to arrange the blocks in consistent
rows or columns so they can utilize a single set of headings.
Another guideline for sketching the layout is to try to arrange the spreadsheet so that it
starts with the data at the top and progresses logically toward the objective cell at the bottom.
This will be easier to understand and follow than if the data cells, changing cells, output cells,
and objective cell are all scattered throughout the spreadsheet.
A sketch of a potential spreadsheet layout for the Everglade problem is shown in Figure 4.2.
The data cells for the interest rates, starting balance, and minimum cash balance are at the
top of the spreadsheet. All of the remaining elements in the spreadsheet then follow the same
structure. The rows represent the different years (from 2018 through 2028). All the various
cash inflows and outflows are then broken out in the columns, starting with the projected cash
flow from the business operations (with data for each of the 10 years), continuing with the
loan inflows, interest payments, and loan paybacks, and culminating with the ending balance
(calculated for each year). The long-term loan is a one-time loan (in 2018), so it is sketched as
a single cell. The short-term loan can occur in any of the 10 years (2018 through 2027), so it
is sketched as a block of cells. The interest payments start one year after the loans. The longterm loan is paid back 10 years later (2028).
Organizing the elements with a consistent structure, like in Figure 4.2, not only saves having to retype the year labels for each element, but also makes the model easier to understand.
Everything that happens in a given year is arranged together in a single row.
FIGURE 4.2
Sketch of the spreadsheet for Everglade’s cash flow problem.
LT Rate
ST Rate
Start Balance
Minimum Cash
Cash
Flow
LT
Loan
ST
Loan
LT
Interest
ST
Interest
LT
Payback
ST
Payback
Ending
Balance
Minimum
Balance
2018
2019
.
.
.
.
2027
2028
≥
4.2 Overview of the Process of Modeling with Spreadsheets 131
It is generally easiest to start sketching the portion of the layout where the data would be
placed. The structure of the rest of the model should then follow the structure of the data cells.
For example, once the projected cash flows data are sketched as a vertical column (with each
year in a row), then it follows that the other cash flows should be structured the same way.
Note that the sketch of the layout shown in Figure 4.2 has a logical progression to the
spreadsheet. The data for the problem are located at the top and left of the spreadsheet. Then,
since the cash flow, loan amounts, interest payments, and loan paybacks are all part of the
calculation for the ending balance, the columns are arranged this way, with the ending balance
directly to the right of all these other elements. Since Sheldon has indicated that the objective
is to maximize the ending balance in 2028, this cell is designated to be the objective cell.
Each year, the balance must be greater than the minimum required balance ($500,000).
Since this will be a constraint in the model, it is logical to arrange the balance and minimum
balance blocks of numbers adjacent to each other in the spreadsheet. You can put the ≥ signs
on the sketch to remind yourself that these will be constraints.
The sketch of a spreadsheet
in Figure 4.2 has a logical
progression, starting with
the data on the top left and
then moving through the
calculations toward the
objective cell on the bottom
right.
Build: Start with a Small Version of the Spreadsheet
Once you’ve thought about a logical layout for the spreadsheet, it is finally time to open a new
worksheet in Excel and start building the model. If it is a complicated model, you may want
to start by building a small, readily manageable version of the model. The idea is to first make
sure that you’ve got the logic of the model worked out correctly for the small version before
expanding the model to full scale.
For example, in the Everglade problem, we could get started by building a model for just
the first two years (2018 and 2019), like the spreadsheet shown in Figure 4.3.
Work out the logic for a
small version of the spreadsheet model before expanding to full size.
FIGURE 4.3
A small version (years 2018 and 2019 only) of the spreadsheet for the Everglade cash flow management problem.
A
1
B
C
D
E
F
G
I
J
ST
Payback
Ending
Balance
H
K
L
Everglade Cash Flow Management Problem (Years 2018 and 2019)
2
3
LT Rate
5%
4
ST Rate
7%
(all cash figures in millions of dollars)
5
6
Start Balance
1
7
Minimum Cash
0.5
10
Year
Cash
Flow
LT
Loan
ST
Loan
11
2018
–8
6
2
12
2019
–2
8
9
LT
Interest
ST
Interest
–0.30
–0.14
5
LT
Payback
–2.00
F
G
H
I
J
9
LT
ST
LT
ST
10
Interest
Interest
Payback
Payback
Ending
Balance
= –LTRate*LTLoan
= –STRate*E11
11
12
Range Name
Cell
LTLoan
LTRate
MinimumCash
StartBalance
STRate
D11
C3
C7
C6
C4
= –E11
Minimum
Balance
1.00
≥
0.50
1.56
≥
0.50
=StartBalance+SUM(C11:I11)
=J11+SUM(C12:I12)
K
L
Minimum
Balance
≥
≥
=MinimumCash
=MinimumCash
132 Chapter Four The Art of Modeling with Spreadsheets
This spreadsheet is set up to follow the layout suggested in the sketch of Figure 4.2. The
loan amounts are in columns D and E. Since the interest payments are not due until the following year, the formulas in columns F and G refer to the loan amounts from the preceding
year (LTLoan, or D11, for the long-term loan, and E11 for the short-term loan). The loan payments are calculated in columns H and I. Column H is blank because the long-term loan does
not need to be repaid until 2028. The short-term loan is repaid one year later, so the formula in
cell I12 refers to the short-term loan taken the preceding year (cell E11). The ending balance
in 2018 is the starting balance plus the sum of all the various cash flows that occur in 2018
(cells C11:I11). The ending balance in 2019 is the ending balance in 2018 (cell J11) plus the
sum of all the various cash flows that occur in 2019 (cells C12:I12). All these formulas are
summarized below the spreadsheet in Figure 4.3.
Building a small version of the spreadsheet works very well for spreadsheets that have
a time dimension. For example, instead of jumping right into a 10-year planning problem,
we can start with the simpler problem of just looking at a couple of years. Once this smaller
model is working correctly, you then can expand the model to 10 years.
Even if a spreadsheet model does not have a time dimension, the same concept of starting
small can be applied. For example, if certain constraints considerably complicate a problem,
start by working on a simpler problem without the difficult constraints. Get the simple model
working and then move on to tackle the difficult constraints. If a model has many sets of output cells, you can build up a model piece by piece by working on one set of output cells at a
time, making sure each set works correctly before moving on to the next.
Test: Test the Small Version of the Model
Try entering numbers in the
changing cells for which
you know what the values
of the output cells should
be.
If you do start with a small version of the model first, be sure to test this version thoroughly to
make sure that all the logic is correct. It is far better to fix a problem early, while the spreadsheet is still a manageable size, rather than later after an error has been propagated throughout
a much larger spreadsheet.
To test the spreadsheet, try entering values in the changing cells for which you know what
the values of the output cells should be, and then see if the spreadsheet gives the results
that you expect. For example, in Figure 4.3, if zeroes are entered for the loan amounts, then
the interest payments and loan payback quantities also should be zero. If $1 million is borrowed for both the long-term loan and the short-term loan, then the interest payments the
following year should be $50,000 and $70,000, respectively. (Recall that the interest rates are
5 percent and 7 percent, respectively.) If Everglade takes out a $6 million long-term loan and a
$2 million short-term loan in 2018, plus a $5 million short-term loan in 2019, then the ending
balances should be $1 million for 2018 and $1.56 million for 2019 (based on the calculations
done earlier by hand). All these tests work correctly for the spreadsheet in Figure 4.3, so we
can be fairly certain that it is correct.
If the output cells are not giving the results that you expect, then carefully look through the
formulas to see if you can determine and fix the problem. Section 4.4 will give further guidance on some ways to debug a spreadsheet model.
Build: Expand the Model to Full-Scale Size
Excel Tip: A shortcut
for filling down or filling
across to the right is to
select the cell you want
to copy, click on the fill
handle (the small box in the
lower right-hand corner of
the selection rectangle), and
drag through the cells you
want to fill.
Once a small version of the spreadsheet has been tested to make sure all the formulas are correct and everything is working properly, the model can be expanded to full-scale size. Excel’s
fill commands often can be used to quickly copy the formulas into the remainder of the model.
To illustrate this process, consider the small version of the Everglade spreadsheet in
Figure 4.3, The formulas in columns F, G, I, J, and L can be copied using the Fill Down command in the Editing Group of the Home tab to obtain all the formulas shown in Figure 4.4.
For example, selecting cells G12:G21 and choosing Fill Down will take the formula in cell
G12 and copy it (while adjusting the cell address in Column E for the formula) into cells G13
through G21. We explain next how this works.
Before using the fill commands to copy formulas, be sure that the relative and absolute
­references have been used appropriately. (Appendix A provides details about relative and
absolute references.) A reference to a range name is an absolute reference. For example, in
G12 (= − STRate*E11), using a range name STRate for the short-term loan interest rate
4.2 Overview of the Process of Modeling with Spreadsheets 133
FIGURE 4.4
A complete spreadsheet model for the Everglade cash flow management problem, including the equations entered into the objective
cell EndBalance (J21) and all the other output cells, to be used before calling on Solver. The entries in the changing cells, LTLoan
(D11) and STLoan (E11:E20), are only a trial solution at this stage.
A
1
B
C
E
D
F
G
H
I
J
K
L
Everglade Cash Flow Management Problem
2
3
LT Rate
5%
4
ST Rate
7%
(all cash figures in millions of dollars)
5
6
Start Balance
1
7
Minimum Cash
0.5
Cash
LT
ST
LT
ST
LT
ST
Ending
10
Year
Flow
Loan
Loan
Interest
Interest
Payback
Payback
Balance
11
2018
–8
6
2
8
9
Minimum
Balance
1.00
≥
0.5
1.56
≥
0.5
12
2019
–2
5
–0.30
–0.14
–2
13
2020
–4
0
–0.30
–0.35
–5
–8.09
≥
0.5
14
2021
3
0
–0.30
0
0
–5.39
≥
0.5
15
2022
6
0
–0.30
0
0
0.31
≥
0.5
16
2023
3
0
–0.30
0
0
3.01
≥
0.5
17
2024
–4
0
–0.30
0
0
–1.29
≥
0.5
18
2025
7
0
–0.30
0
0
5.41
≥
0.5
19
2026
–2
0
0
3.11
≥
0.5
2027
10
0
–0.30
–0.30
0
20
0
0
12.81
≥
0.5
21
2028
–0.30
0
0
6.51
≥
0.5
F
–6
G
H
I
J
9
LT
ST
LT
ST
Ending
10
Interest
Interest
Payback
Payback
11
Balance
K
L
Minimum
Balance
=StartBalance+SUM(C11:I11)
≥ =MinimumCash
12
=–LTRate*LTLoan
=–STRate*E11
=–E11
=J11+SUM(C12:I12)
≥ =MinimumCash
13
=–LTRate*LTLoan
=–STRate*E12
=–E12
=J12+SUM(C13:I13)
≥ =MinimumCash
14
=–LTRate*LTLoan
=–STRate*E13
=–E13
=J13+SUM(C14:I14)
≥ =MinimumCash
15
=–LTRate*LTLoan
=–STRate*E14
=–E14
=J14+SUM(C15:I15)
≥ =MinimumCash
16
=–LTRate*LTLoan
=–STRate*E15
=–E15
=J15+SUM(C16:I16)
≥ =MinimumCash
17
=–LTRate*LTLoan
=–STRate*E16
=–E16
=J16+SUM(C17:I17)
≥ =MinimumCash
18
=–LTRate*LTLoan
=–STRate*E17
=–E17
=J17+SUM(C18:I18)
≥ =MinimumCash
19
=–STRate*E18
=–E18
=J18+SUM(C19:I19)
≥ =MinimumCash
20
=–LTRate*LTLoan
=–LTRate*LTLoan
=–STRate*E19
=–E19
=J19+SUM(C20:I20)
≥ =MinimumCash
21
=–LTRate*LTLoan
=–STRate*E20
=–LTLoan =–E20
=J20+SUM(C21:I21)
≥ =MinimumCash
Range Name
Cells
CashFlow
EndBalance
EndingBalance
LTLoan
LTRate
MinimumBalance
MinimumCash
StartBalance
STLoan
STRate
C11:C20
J21
J11:J21
D11
C3
L11:L21
C7
C6
E11:E20
C4
134 Chapter Four The Art of Modeling with Spreadsheets
Excel Tip: A shortcut for
changing a cell reference
from relative to absolute is
to press the F4 key on a PC
or command-T on a Mac.
makes this an absolute reference. When copied into cells G13:G21, the short-term loan interest rate used will always be the value in STRate (C4).
By contrast, a reference to a specific cell address is a relative reference. For example, the
reference to E11 (the loan amount from the previous year) in the formula for G12 is a relative
reference. E11 is two cells to the left and one cell up. When the formula is copied from G12
into G13:G21, the reference in each of these cells will continue to be two cells to the left and
one cell up. This is exactly what we want, since we always want the interest payment to be
based on the short-term loan that was taken one year ago (two cells to the left and one cell up).
After using the Fill Down command to copy the formulas in columns F, G, I, J, and L
and entering the LT loan payback into cell H21, the complete model appears as shown in
Figure 4.4.
Test: Test the Full-Scale Version of the Model
Just as it was important to test the small version of the model, it needs to be tested again after
it is expanded to full-scale size. The procedure is the same one followed for testing the small
version, including the ideas that will be presented in Section 4.4 for debugging a spreadsheet
model.
Analyze: Analyze the Model
Before using Solver, the spreadsheet in Figure 4.4 is merely an evaluative model for Everglade. It can be used to evaluate any proposed solution, including quickly determining what
interest and loan payments will be required and what the resulting balances will be at the end
of each year. For example, LTLoan (D11) and STLoan (E11:E20) in Figure 4.4 show one
­possible plan, which turns out to be unacceptable because EndingBalance (J11:J21) indicates
that a negative ending balance would result in three of the years.
To optimize the model, Solver is used as shown in Figure 4.5 to specify the objective cell,
the changing cells, and the constraints. Everglade management wants to find a combination of
loans that will keep the company solvent throughout the next 10 years (2018–2027) and then
will leave as large a cash balance as possible in 2028 after paying off all the loans. Therefore,
the objective cell to be maximized is EndBalance (J21) and the changing cells are the loan
amounts LTLoan (D11) and STLoan (E11:E20). To assure that Everglade maintains a minimum balance of at least $500,000 at the end of each year, the constraints for the model are
EndingBalance (J11:J21) ≥ MinimumBalance (L11:L21).
After running Solver, the optimal solution is shown in Figure 4.5. The changing cells,
LTLoan (D11) and STLoan (E11:E20), give the loan amounts in the various years. The objective cell EndBalance (J21) indicates that the ending balance in 2028 will be $5.39 million.
Conclusion of the Case Study
The spreadsheet model developed by Everglade’s CFO, Julie Lee, is the one shown in
­Figure 4.5. Her next step is to submit a report to her CEO, Sheldon Lee, that recommends the
plan obtained by this model.
Soon thereafter, Sheldon and Julie meet to discuss her report.
Sheldon: Thanks for your report, Julie. Excellent job. Your spreadsheet really lays everything out in a very understandable way.
Julie: Thanks. It took a little while to get the spreadsheet organized properly and to make
sure it was operating correctly, but I think the time spent was worthwhile.
Sheldon: Yes, it was. You can’t rush those things. But one thing is still bothering me.
Julie: What’s that?
Sheldon: It has to do with our forecasts for the company’s future cash flows. We have been
assuming that the cash flows in the coming years will be the ones shown in column C of your
spreadsheet. Those are good estimates, but we both know that they are only estimates. A lot of
changes that we can’t foresee now are likely to occur over the next 10 years. When there is a
shift in the economy, or when other unexpected developments occur that impact the company,
those cash flows can change a lot. How do we know if your recommended plan will still be a
good one if those kinds of changes occur?
4.2 Overview of the Process of Modeling with Spreadsheets 135
FIGURE 4.5
A complete spreadsheet model for the Everglade cash flow management problem after calling on Solver to obtain the optimal solution shown in the changing cells, LTLoan (D11) and STLoan (E11:E20). The objective cell EndBalance (J21) indicates that the
resulting cash balance in 2028 will be $5.39 million if all the data cells prove to be accurate.
A
1
B
C
E
D
F
G
H
I
J
K
L
Everglade Cash Flow Management Problem
2
3
LT Rate
5%
4
ST Rate
7%
(all cash figures in millions of dollars)
5
6
Start Balance
1
7
Minimum Cash
0.5
Cash
LT
ST
LT
ST
LT
ST
Ending
10
Year
Flow
Loan
Loan
Interest
Interest
Payback
Payback
Balance
11
2018
–8
4.65
2.85
12
2019
–2
5.28
–0.23
–0.20
13
2020
–4
9.88
–0.23
14
2021
3
7.81
–0.23
15
2022
6
2.59
–0.23
16
2023
3
0
17
2024
–4
18
2025
19
8
9
Minimum
Balance
0.50
≥
–2.85
0.50
≥
0.50
–0.37
–5.28
0.50
–9.88
0.50
–0.55
–7.81
0.50
≥
≥
≥
0.50
–0.69
–0.23
–0.18
–2.59
0.50
≥
0.50
4.23
–0.23
0
0
0.50
≥
0.50
7
0
–0.23
–0.30
– 4.23
2.74
≥
0.50
2026
–2
0
0
0
0.51
2027
10
0
0
0
10.27
≥
≥
0.50
20
–0.23
–0.23
21
2028
–0.23
0
0
5.39
≥
0.50
–4.65
F
G
H
I
J
9
LT
ST
LT
ST
Ending
10
Interest
Interest
Payback
Payback
11
0.50
0.50
0.50
0.50
K
L
Minimum
Balance
Balance
= StartBalance+SUM(C11:I11)
≥ =MinimumCash
12
=–LTRate*LTLoan
=–STRate*E11
=–E11
=J11+SUM(C12:I12)
≥ =MinimumCash
13
=–LTRate*LTLoan
=–STRate*E12
=–E12
=J12+SUM(C13:I13)
14
=–LTRate*LTLoan
=–STRate*E13
=–E13
=J13+SUM(C14:I14)
15
=–LTRate*LTLoan
=–STRate*E14
=–E14
=J14+SUM(C15:I15)
≥ =MinimumCash
≥ =MinimumCash
≥ =MinimumCash
16
=–LTRate*LTLoan
=–STRate*E15
=–E15
=J15+SUM(C16:I16)
≥ =MinimumCash
17
=–LTRate*LTLoan
=–STRate*E16
=–E16
=J16+SUM(C17:I17)
≥ =MinimumCash
18
=–LTRate*LTLoan
=–STRate*E17
=–E17
=J17+SUM(C18:I18)
19
≥ =MinimumCash
=–STRate*E18
=–E18
=J18+SUM(C19:I19)
20
=–LTRate*LTLoan
=–LTRate*LTLoan
=–STRate*E19
=–E19
=J19+SUM(C20:I20)
≥ =MinimumCash
≥ =MinimumCash
21
=–LTRate*LTLoan
=–STRate*E20
=–LTLoan =–E20
=J20+SUM(C21:I21)
≥ =MinimumCash
Solver Parameters
Set Objective Cell: EndBalance
To: Max
By Changing Variable Cells:
LTLoan, STLoan
Subject to the Constraints:
EndingBalance >= MinimumBalance
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
Range Name
Cells
CashFlow
EndBalance
EndingBalance
LTLoan
LTRate
MinimumBalance
MinimumCash
StartBalance
STLoan
STRate
C11:C20
J21
J11:J21
D11
C3
L11:L21
C7
C6
E11:E20
C4
136 Chapter Four The Art of Modeling with Spreadsheets
Julie: A very good question. To answer it, we should do some what-if analysis to see what
would happen if those kinds of changes occur. Now that the spreadsheet is set up properly,
it will be easy to do that by simply changing some of the cash flows in column C and seeing
what would happen with the current plan. You can try out any change or changes you want
and immediately see the effect. Each time you change a future cash flow, you also have the
option of trying out changes on short-term loan amounts to see what kind of adjustments
would be needed to maintain a balance of at least $500,000 in every year.
OK, are you ready? Shall we do some what-if analysis now?
Sheldon: Let’s do.
Providing a data cell for
each piece of data makes it
easy to check what would
happen if the correct value
for a piece of data differs
from its initial estimate.
Fortunately, Julie had set up the spreadsheet properly (providing a data cell for the cash
flow in each of the next 10 years) to enable performing what-if analysis immediately by simply trying different numbers in some of these data cells. (The next chapter will focus on
describing the importance of what-if analysis and alternative ways of performing this kind of
analysis.) After spending half an hour trying different numbers, Sheldon and Julie conclude
that the plan in Figure 4.5 will be a sound initial financial plan for the next 10 years, even
if future cash flows deviate somewhat from current forecasts. If deviations do occur, adjustments will of course need to be made in the short-term loan amounts. At any point, Julie also
will have the option of returning to the company’s bank to try to arrange another long-term
loan for the remainder of the 10 years at a lower interest rate than that offered for short-term
loans. If so, essentially the same spreadsheet model as in Figure 4.5 can be used, along with
Solver, to find the optimal adjusted financial plan for the remainder of the 10 years.
A management science technique called computer simulation provides another effective way
of taking the uncertainty of future cash flows into account. Chapters 12 and 13 will describe this
technique and then Problem 13.24 will continue the analysis of this same case study.
Review
Questions
1. What is a good way to get started with a spreadsheet model if you don’t even know where to
begin?
2. What are two ways in which doing calculations by hand can help you?
3. Describe a useful way to get started organizing and laying out a spreadsheet.
4. What types of values should be put into the changing cells to test the model?
5. What is the difference between an absolute cell reference and a relative cell reference?
4.3
SOME GUIDELINES FOR BUILDING “GOOD” SPREADSHEET MODELS
There are many ways to set up a model on a spreadsheet. While one of the benefits of spreadsheets is the flexibility they offer, this flexibility also can be dangerous. Although Excel provides many features (such as range names, shading, borders, etc.) that allow you to create
“good” spreadsheet models that are easy to understand, easy to debug, and easy to modify,
it is also easy to create “bad” spreadsheet models that are difficult to understand, difficult to
debug, and difficult to modify. The goal of this section is to provide some guidelines that will
help you to create “good” spreadsheet models.
Enter the Data First
All the data should be laid
out on the spreadsheet
before beginning to formulate the rest of the spreadsheet model.
Any spreadsheet model is driven by the data in the spreadsheet. The form of the entire model
is built around the structure of the data. Therefore, it is always a good idea to enter and carefully lay out all the data before you begin to set up the rest of the model. The model structure
then can conform to the layout of the data as closely as possible.
The reason for entering the data first is that it is easier to set up the rest of the model when
the data are already on the spreadsheet. In the Everglade problem (see Figure 4.5), the data for
the cash flows have been laid out in the first columns of the spreadsheet (B and C), with the
year labels in column B and the data in cells C11:C20. Once the data are in place, the layout
for the rest of the model quickly falls into place around the structure of the data. It is only
logical to lay out the changing cells and output cells using the same structure, with each of the
various cash flows in columns that utilize the same row labels from column B.
An Application Vignette
Welch’s, Inc., is the world’s largest processor of Concord and
Niagara grapes with large annual sales that reached $608 million
in 2013. Such products as Welch’s grape jelly and Welch’s grape
juice have been enjoyed by generations of American consumers.
Every September, growers begin delivering grapes to processing plants that then press the raw grapes into juice. Time
must pass before the grape juice is ready for conversion into
finished jams, jellies, juices, and concentrates.
Deciding how to use the grape crop is a complex task given
changing demand and uncertain crop quality and quantity.
Typical decisions include what recipes to use for major product groups, the transfer of grape juice between plants, and the
mode of transportation for these transfers.
Because Welch’s lacked a formal system for optimizing raw
material movement and the recipes used for production, a
management science team developed a preliminary linear programming model. This was a large model with 8,000 decision
variables that focused on the component level of detail. Smallscale testing proved that the model worked.
To make the model more useful, the team then revised it by
aggregating demand by product group rather than by component. This reduced its size to 324 decision variables and 361
functional constraints. The model then was incorporated into
a spreadsheet.
The company has run the continually updated version of
this spreadsheet model each month to provide senior management with information on the optimal logistics plan generated by Solver. The savings from using and optimizing this
model were approximately $150,000 in the first year alone.
A major advantage of incorporating the linear programming
model into a spreadsheet has been the ease of explaining
the model to managers with differing levels of mathematical
understanding. This has led to a widespread appreciation of
the management science approach for both this application
and others.
Source: E. W. Schuster and S. J. Allen, “Raw Material Management at
Welch’s, Inc.,” Interfaces 28, no. 5 (September—October 1998), pp. 13–24.
(A link to this article is available at www.mhhe.com/Hillier6e.)
Now reconsider the spreadsheet model developed in Section 2.2 for the Wyndor Glass Co.
problem. The spreadsheet is repeated here in Figure 4.6. The data for the Hours Used per Unit
Produced have been laid out in the center of the spreadsheet in cells C7:D9. The output cells,
HoursUsed (E7:E9), then have been placed immediately to the right of these data and to the
left of the data on HoursAvailable(G7:G9), where the row labels for these output cells are the
same as for all these data. This makes it easy to interpret the three constraints being laid out
in rows 7–9 of the spreadsheet model. Next, the changing cells and objective cell have been
placed together in row 12 below the data, where the column labels for the changing cells are
the same as for the columns of data above.
The locations of the data occasionally will need to be shifted somewhat to better accommodate the overall model. However, with this caveat, the model structure generally should
conform to the data as closely as possible.
Organize and Clearly Identify the Data
Provide labels in the
spreadsheet that clearly
identify all the data.
Related data should be grouped together in a convenient format and entered into the spreadsheet with labels that clearly identify the data. For data that are laid out in tabular form, the
table should have a heading that provides a general description of the data and then each row
and column should have a label that will identify each entry in the table. The units of the data
also should be identified. Different types of data should be well separated in the spreadsheet.
However, if two tables need to use the same labels for either their rows or columns, then be
consistent in making them either rows in both tables or columns in both tables.
In the Wyndor Glass Co. problem (Figure 4.6), the three sets of data have been grouped
into tables and clearly labeled Unit Profit, Hours Used per Unit Produced, and Hours Available. The units of the data are identified (dollar signs are included in the unit profit data
and hours are indicated in the labels of the time data). Finally, all three data tables make
consistent use of rows and columns. Since the Unit Profit data have their product labels
(Doors and Windows) in columns C and D, the Hours Used per Unit Produced data use this
same structure.
This structure also is carried through to the changing cells (Units Produced). Similarly,
the data for each plant (rows 7–9) are in the rows for both the Hours Used per Unit ­Produced
data and the Hours Available data. Keeping the data oriented the same way is not only less
confusing, but it also makes it possible to use the SUMPRODUCT function. Recall that the
SUMPRODUCT function assumes that the two ranges are exactly the same shape (i.e., the
same number of rows and the same number of columns). If the Unit Profit data and the Units
137
138 Chapter Four The Art of Modeling with Spreadsheets
FIGURE 4.6
The spreadsheet model
formulated in Section 2.2
for the Wyndor Glass Co.
product-mix problem.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Used
Hours
Available
7
Plant 1
1
0
2
≤
4
8
Plant 2
0
2
12
≤
12
9
Plant 3
3
2
18
≤
18
Doors
Windows
Total Profit
2
6
$3,600
10
11
12
Units Produced
Solver Parameters
Set Objective Cell: TotalProfit
To: Max
By Changing Variable Cells:
UnitsProduced
Subject to the Constraints:
HoursUsed <= HoursAvailable
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
E
5
Hours
6
7
8
9
Used
=SUMPRODUCT(C7:D7,UnitsProduced)
=SUMPRODUCT(C8:D8,UnitsProduced)
=SUMPRODUCT(C9:D9,UnitsProduced)
G
11
12
Total Profit
=SUMPRODUCT(Unit Profit,UnitsProduced)
Range Name
Cells
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
TotalProfit
UnitProfit
UnitsProduced
G7:G9
E7:E9
C7:D9
G12
C4:D4
C12:D12
Produced data had not been oriented the same way (e.g., one in a column and the other in a
row), it would not have been possible to use the SUMPRODUCT function in the Total Profit
calculation.
Similarly, for the Everglade problem (Figure 4.5), the five sets of data have been grouped into
cells and tables and clearly labeled ST Rate, LT Rate, Start Balance, Cash Flow, and Minimum
Cash. The units of the data are identified (cells F5:I5 specify that all cash figures are in millions
of dollars), and all the tables make consistent use of rows and columns (years in the rows).
Enter Each Piece of Data into One Cell Only
Every formula using the
same piece of data should
refer to the same single
data cell.
If a piece of data is needed in more than one formula, then refer to the original data cell rather
than repeating the data in additional places. This makes the model much easier to modify. If
the value of that piece of data changes, it only needs to be changed in one place. You do not
need to search through the entire model to find all the places where the data value appears.
For example, in the Everglade problem (Figure 4.5), there is a company policy of maintaining a cash balance of at least $500,000 at all times. This translates into a constraint for
the minimum balance of $500,000 at the end of each year. Rather than entering the minimum
cash position of 0.5 (in millions of dollars) into all the cells in column L, it is entered once in
MinimumCash (C7) and then referred to by the cells in MinimumBalance (L11:L21). Then,
if this policy were to change to, say, a minimum of $200,000 cash, the number would need to
be changed in only one place.
4.3 Some Guidelines for Building “Good” Spreadsheet Models 139
Separate Data from Formulas
Formulas should refer to
data cells for any needed
numbers.
Avoid using numbers directly in formulas. Instead, enter any needed numbers into data cells
and then refer to the data cells as needed. For example, in the Everglade problem (Figure 4.5),
all the data (the interest rates, starting balance, minimum cash, and projected cash flows) are
entered into separate data cells on the spreadsheet. When these numbers are needed to calculate the interest charges (in columns F and G), loan payments (in column H and I), ending
balances (column J), and minimum balances (column L), the data cells are referred to rather
than entering these numbers directly in the formulas.
Separating the data from the formulas has a couple advantages. First, all the data are visible on the spreadsheet rather than buried in formulas. Seeing all the data makes the model
easier to interpret. Second, the model is easier to modify since changing data only requires
modifying the corresponding data cells. You don’t need to modify any formulas. This proves
to be very important when it comes time to perform what-if analysis (the subject of the next
chapter) to see what the effect would be if some of the estimates in the data cells were to take
on other plausible values.
Keep It Simple
Make the spreadsheet as
easy to interpret as possible.
Avoid the use of powerful Excel functions when simpler functions are available that are easier
to interpret. As much as possible, stick to SUMPRODUCT or SUM functions. This makes
the model easier to understand and also helps to ensure that the model will be linear. (Linear
models are considerably easier to solve than others.) Try to keep formulas short and simple. If
a complicated formula is required, break it out into intermediate calculations with subtotals.
For example, in the Everglade spreadsheet, each element of the loan payments is broken out
explicitly: LT Interest, ST Interest, LT Payback, and ST Payback. Some of these columns
could have been combined (e.g., into two columns with LT Payments and ST Payments, or
even into one column for all Loan Payments). However, this makes the formulas more complicated and also makes the model harder to test and debug. As laid out, the individual formulas
for the loan payments are so simple that their values can be predicted easily without even
looking at the formula. This simplifies the testing and debugging of the model.
Use Range Names
Range names make formulas much easier to interpret.
One way to refer to a block of related cells (or even a single cell) in a spreadsheet formula is
to use its cell address (e.g., L11:L21 or C3). However, when reading the formula, this requires
looking at that part of the spreadsheet to see what kind of information is given there. As
mentioned previously in Sections 1.2 and 2.2, a better alternative is to assign a descriptive
range name to the block of cells that immediately identifies what is there. (This is done by
selecting the block of cells, clicking on the name box on the left of the formula bar above the
spreadsheet, and then typing a name.) This is especially helpful when writing a formula for
an output cell. Writing the formula in terms of range names instead of cell addresses makes
the formula much easier to interpret. Range names also make the description of the model in
Solver much easier to understand.
Figure 4.5 illustrates the use of range names for the Everglade spreadsheet model. For
example, consider the formula for long-term interest in cell F12. Since the long-term rate
is given in cell C3 and the long-term loan amount is in cell D11, the formula for the longterm interest could have been written as = − C3*D11. However, by using the range name
LTRate for cell C3 and the range name LTLoan for cell D11, the formula instead becomes =
− LTRate*LTLoan, which is much easier to interpret at a glance.
On the other hand, be aware that it is easy to get carried away with defining range names.
Defining too many range names can be more trouble than it is worth. For example, when
related data are grouped together in a table, we recommend giving a range name only for the
entire table rather than for the individual rows and columns. In general, we suggest defining
range names only for each group of data cells, the changing cells, the objective cell, and both
sides of each group of constraints (the left-hand side and the right-hand side).
Care also should be taken to assure that it is easy to quickly identify which cells are
referred to by a particular range name. Use a name that corresponds exactly to the label on
the spreadsheet. For example, in Figure 4.5, columns J and L are labeled Ending Balance
140 Chapter Four The Art of Modeling with Spreadsheets
Spaces are not allowed in
range names. When a range
name has more than one
word, we have used capital
letters to distinguish the start
of each new word in a range
name (e.g., MinimumBalance). Another way is to use
the underscore character
(e.g., Minimum_Balance).
and Minimum Balance on the spreadsheet, so we use the range names EndingBalance and
Minimum­Balance. Using exactly the same name as the label on the spreadsheet makes it
quick and easy to find the cells that are referred to by a range name.
When desired, a list of all the range names and their corresponding cell addresses can be
pasted directly into the spreadsheet by choosing Paste from the Use in Formula menu on the
Formulas tab, and then clicking Paste List. Such a list (after reformatting) is included below
essentially all the spreadsheets displayed in this text.
When modifying an existing model that utilizes range names, care should be taken to assure
that the range names continue to refer to the correct range of cells. When inserting a row or
column into a spreadsheet model, it is helpful to insert the row or column into the middle of
a range rather than at the end. For example, to add another product to a product-mix model
with four products, add a column between Products 2 and 3 rather than after Product 4. This
will automatically extend the relevant range names to span across all five columns since these
range names will continue to refer to everything between Product 1 and Product 4, including
the newly inserted column for the fifth product. Similarly, deleting a row or column from the
middle of a range will contract the span of the relevant range names appropriately. You can
double-check the cells that are referred to by a range name by choosing that range name from
the name box (on the left of the formula bar above the spreadsheet). This will highlight the
cells that are referred to by the chosen range name.
Use Relative and Absolute References to Simplify Copying Formulas
Excel’s fill commands
provide a quick and reliable
way to replicate a formula
into multiple cells.
Whenever multiple related formulas will be needed, try to enter the formula just once and
then use Excel’s fill commands to replicate the formula. Not only is this quicker than retyping
the formula, but it is also less prone to error.
We saw a good example of this when discussing the expansion of the model to full-scale
size in the preceding section. Starting with the two-year spreadsheet in Figure 4.3, fill commands were used to copy the formulas in columns F, G, I, J, and L for the remaining years to
create the full-scale, 10-year spreadsheet in Figure 4.4.
Using relative and absolute references for related formulas not only aids in building a
model but also makes it easier to modify an existing model or template. For example, suppose
that you have formulated a spreadsheet model for a product-mix problem but now wish to
modify the model to add another resource and its constraint. This requires inserting a row into
the spreadsheet. If the output cells are written with proper relative and absolute references,
then it is simple to copy the existing formulas into the inserted row.
Use Borders, Shading, and Colors to Distinguish between Cell Types
Make it easy to spot all the
cells of the same type.
It is important to be able to easily distinguish between the data cells, changing cells, output
cells, and objective cell in a spreadsheet. One way to do this is to use different borders and cell
shading for each of these different types of cells. In the text, data cells appear lightly shaded,
changing cells are shaded a medium amount with a light border, output cells appear with no
shading, and the objective cell is shaded darkly with a heavy border.
In the spreadsheet files for this chapter, data cells are light blue, changing cells are yellow, and the objective cell is orange. Obviously, you may use any scheme that you like. The
important thing is to be consistent, so that you can quickly recognize the types of cells.
Then, when you want to examine the cells of a certain type, the color will immediately
guide you there.
Show the Entire Model on the Spreadsheet
Solver uses a combination of the spreadsheet and the Solver dialog box (or the model pane in
Analytic Solver) to specify the model to be solved. Therefore, it is possible to include certain
elements of the model (such as the ≤, =, or ≥ signs and/or the right-hand sides of the constraints) in Solver without displaying them in the spreadsheet. However, we strongly recommend that every element of the model be displayed on the spreadsheet. Every person using or
adapting the model, or referring back to it later, needs to be able to interpret the model. This is
much easier to do by viewing the model on the spreadsheet than by trying to decipher it from
Solver. Furthermore, a printout of the spreadsheet does not include information from Solver.
4.3 Some Guidelines for Building “Good” Spreadsheet Models 141
Display every element of
the model on the spreadsheet rather than relying
on only Solver to include
certain elements.
In particular, all the elements of a constraint should be displayed on the spreadsheet. For
each constraint, three adjacent cells should be used for the total of the left-hand side, the ≤,
=, or ≥ sign in the middle, and the right-hand side. (Note in Figure 4.5 that this was done in
columns J, K, and L of the spreadsheet for the Everglade problem.) As mentioned earlier, the
changing cells and objective cell should be highlighted in some manner (e.g., with borders
and/or cell shading). A good test is that you should not need to go to Solver to determine any
element of the model. You should be able to identify the changing cells, the objective cell, and
all the constraints in the model just by looking at the spreadsheet.
A Poor Spreadsheet Model
It is certainly possible to set up a linear programming spreadsheet model without utilizing
any of these ideas. Figure 4.7 shows an alternative spreadsheet formulation for the Everglade
problem that violates nearly every one of these guidelines. This formulation can still be solved
using Solver, which in fact yields the same optimal solution as in Figure 4.5. However, the
formulation has many problems. It is not clear which cells yield the solution (borders and/
or shading are not used to highlight the changing cells and objective cell). Without going to
Solver, the constraints in the model cannot be identified (the spreadsheet does not show the
entire model). The spreadsheet also does not show most of the data. For example, to determine the data used for the projected cash flows, the interest rates, or the starting balance, you
FIGURE 4.7
A poor formulation of the
spreadsheet model for the
Everglade cash flow management problem.
A
1
B
C
D
E
A Poor Formulation of the Everglade Cash Flow Problem
2
LT
ST
Ending
4
Year
Loan
Loan
Balance
5
2018
4.65
2.85
0.50
6
2019
5.28
0.50
7
2020
9.88
0.50
8
2021
7.81
0.50
9
2022
2.59
0.50
10
2023
0
0.50
11
2024
4.23
0.50
12
2025
0
2.74
13
2026
0
0.51
14
2027
0
10.27
15
2028
3
Solver Parameters
Set Objective Cell: E15
To: Max
By Changing Variable Cells:
C5, D5:D14
Subject to the Constraints:
E5:E15 >= 0.5
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
5.39
E
3
Ending
4
Balance
5
6
=1–8+C5+D5
7
=E6–4+D7–$C$5*(0.05)–D6*(1.07)
8
=E7+3+D8–$C$5*(0.05)–D7*(1.07)
9
=E8+6+D9–$C$5*(0.05)–D8*(1.07)
10
=E9+3+D10–$C$5*(0.05)–D9*(1.07)
11
=E10–4+D11–$C$5*(0.05)–D10*(1.07)
12
=E11+7+D12–$C$5*(0.05)–D11*(1.07)
13
=E12–2+D13–$C$5*(0.05)–D12*(1.07)
14
=E13+10+D14–$C$5*(0.05)–D13*(1.07)
15
=E14+D15–$C$5*(1.05)–D14*(1.07)
=E5–2+D6–$C$5*(0.05)–D5*(1.07)
F
142 Chapter Four The Art of Modeling with Spreadsheets
need to dig into the formulas in column E (the data are not separate from the formulas). If any
of these data change, the actual formulas need to be modified rather than simply changing a
number on the spreadsheet. Furthermore, the formulas and the model in Solver are difficult to
interpret (range names are not utilized).
Compare Figures 4.5 and 4.7. Applying the guidelines for good spreadsheet models (as
is done for Figure 4.5) results in a model that is easier to understand, easier to debug, and
easier to modify. This is especially important for models that will have a long life span. If this
model is going to be reused months later, the “good” model of Figure 4.5 immediately can be
understood, modified, and reapplied as needed, whereas deciphering the spreadsheet model
of Figure 4.7 again would be a great challenge.
Review
Questions
4.4
1. Which part of the model should be entered first on the spreadsheet?
2. Should numbers be included in formulas or entered separately in data cells?
3. How do range names make formulas and the model in Solver easier to interpret? How should
range names be chosen?
4. What are some ways to distinguish data cells, changing cells, output cells, and objective cells
on a spreadsheet?
5. How many cells are needed to completely specify a constraint on a spreadsheet?
DEBUGGING A SPREADSHEET MODEL
Debugging a spreadsheet
model sometimes is as
challenging as debugging a
computer program.
If you have added rows or
columns to the spreadsheet,
make sure that each of the
range names still refers to
the correct cells.
Excel Tip: Pressing
­control-~ on a PC (or
­command-~ on a Mac)
toggles the worksheet
between viewing values
and viewing formulas in
all the output cells.
Excel’s auditing tools
enable you to either trace
forward or backward to see
the linkages between cells.
No matter how carefully it is planned and built, even a moderately complicated model usually
will not be error-free the first time it is run. Often the mistakes are immediately obvious and
quickly corrected. However, sometimes an error is harder to root out. Following the guidelines in Section 4.3 for developing a good spreadsheet model can make the model much easier
to debug. Even so, much like debugging a computer program, debugging a spreadsheet model
can be a difficult task. This section presents some tips and a variety of Excel features that can
make debugging easier.
As a first step in debugging a spreadsheet model, test the model using the principles discussed in the first subsection on testing in Section 4.2. In particular, try different values for
the changing cells for which you can predict the correct result in the output cells and see if
they calculate as expected. Values of 0 are good ones to try initially because usually it is then
obvious what should be in the output cells. Try other simple values, such as all 1s, where the
correct results in the output cells are reasonably obvious. For more complicated values, break
out a calculator and do some manual calculations to check the various output cells. Include
some very large values for the changing cells to ensure that the calculations are behaving reasonably for these extreme cases.
If you have defined range names, be sure that they still refer to the correct cells. Sometimes they can become disjointed when you add rows or columns to the spreadsheet. To test
the range names, you can either select the various range names in the name box, which will
highlight the selected range in the spreadsheet, or paste the entire list of range names and their
references into the spreadsheet.
Carefully study each formula to be sure it is entered correctly. A very useful feature in
Excel for checking formulas is the toggle to switch back and forth between viewing the formulas in the worksheet and viewing the resulting values in the output cells. By default, Excel
shows the values that are calculated by the various output cells in the model. Typing control-~
switches the current worksheet to instead display the formulas in the output cells, as shown in
Figure 4.8. Typing control-~ again switches back to the standard view of displaying the values
in the output cells (like Figure 4.5).
Another useful set of features built into Excel are the auditing tools. The auditing tools
are available in the Formula Auditing group of the Formulas Tab.
The auditing tools can be used to graphically display which cells make direct links to a
given cell. For example, selecting LTLoan (D11) in Figure 4.5 and then Trace Dependents
generates the arrows on the spreadsheet shown in Figure 4.9.
LT Rate
ST Rate
4
–2
–4
3
3
–4
7
–2
10
Year
2018
2019
2020
2021
2022
2023
2024
2025
2026
2027
2028
10
D
E
11
12
13
14
15
16
17
18
19
20
21
LT
ST
6
G
ST
=–STRate*E15
=–LTRate*LTLoan
0
0
0
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
=–LTRate*LTLoan
4.23256 =–LTRate*LTLoan
=–STRate*E20
=–STRate*E18
=–STRate*E19
=–STRate*E17
=–STRate*E16
=–STRate*E14
2.58639 =–LTRate*LTLoan
0
=–STRate*E13
7.80732 =–LTRate*LTLoan
=–STRate*E12
9.88295 =–LTRate*LTLoan
Interest
=–STRate*E11
Interest
LT
(all cash figures in millions of dollars)
F
5.28073 =–LTRate*LTLoan
Flow Loan
Loan
–8 4.65124 2.84759
Cash
0.5
Start Balance
Minimum Cash
7
8
9
1
0.07
0.05
6
5
C
Everglade Cash Flow Management Problem
B
2
3
1
A
=–E18
=–E19
=–E17
=–E16
=–E15
=–E14
=–E13
=–E12
=–E11
Payback
ST
I
=–LTLoan =–E20
Payback
LT
H
=J20+SUM(C21:I21)
=J18+SUM(C19:I19)
=J19+SUM(C20:I20)
=J17+SUM(C18:I18)
=J16+SUM(C17:I17)
=J15+SUM(C16:I16)
=J14+SUM(C15:I15)
=J13+SUM(C14:I14)
=J12+SUM(C13:I13)
=J11+SUM(C12:I12)
Balance
=StartBalance+SUM(C11:I11)
Ending
J
≥
≥
≥
≥
≥
≥
≥
≥
≥
≥
≥
K
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
=MinimumCash
Balance
=MinimumCash
Minimum
L
The spreadsheet obtained by toggling the spreadsheet in Figure 4.5 once to replace the values in the output cells by the formulas entered into those cells. Using the toggle
feature in Excel once more will restore the view of the spreadsheet shown in Figure 4.5.
FIGURE 4.8
4.4 Debugging a Spreadsheet Model 143
144 Chapter Four The Art of Modeling with Spreadsheets
FIGURE 4.9
The spreadsheet obtained by using the Excel auditing tools to trace the dependents of the LT Loan value in cell D11
of the spreadsheet in Figure 4.5.
A
1
B
C
D
E
F
G
H
I
J
K
L
Everglade Cash Flow Management Problem
2
3
LT Rate
5%
4
ST Rate
7%
(all cash figures in millions of dollars)
5
6
Start Balance
1
7
Minimum Cash
0.5
Cash
LT
ST
LT
ST
LT
ST
Ending
10
Year
Flow
Loan
Loan
Interest
Interest
Payback
Payback
Balance
11
2018
–8
4.65
2.85
8
9
Minimum
Balance
0.50
≥
0.5
–2
5.28
–0.23
–0.20
–2.85
0.50
≥
0.5
–4
9.88
–0.23
–0.37
–5.28
0.50
≥
0.5
2021
3
7.81
–0.23
–0.69
–9.88
0.50
≥
0.5
15
2022
6
2.59
–0.23
–0.55
–7.81
0.50
≥
0.5
16
2023
3
0
–0.23
–0.18
–2.59
0.50
≥
0.5
0.5
12
2019
13
2020
14
17
2024
–4
4.23
–0.23
0
0
0.50
≥
18
2025
7
0
–0.23
–0.30
–4.23
2.74
≥
0.5
19
2026
–2
0
0.5
20
2027
10
0
21
2028
The Evaluate Formula
auditing tool can be used to
evaluate a formula one step
at a time. (At press time,
this particular auditing tool
is only available in Windows versions of Excel.)
–0.23
–0.23
0
0
0.51
≥
0
0
10.27
≥
0.5
–0.23
0
0
5.39
≥
0.5
–4.65
You now can immediately see that LTLoan (D11) is used in the calculation of LT Interest for every year in column F, in the calculation of LTPayback (H21), and in the calculation
of the ending balance in 2018 (J11). This can be very illuminating. Think about what output
cells LTLoan should impact directly. There should be an arrow to each of these cells. If, for
example, LTLoan is missing from any of the formulas in column F, the error will be immediately revealed by the missing arrow. Similarly, if LTLoan is mistakenly entered in any of the
short-term loan output cells, this will show up as extra arrows.
You also can trace backward to see which cells provide the data for any given cell. These
can be displayed graphically by choosing Trace Precedents. For example, choosing Trace
Precedents for the ST Interest cell for 2019 (G12) displays the arrows shown in Figure 4.10.
These arrows indicate that the ST Interest cell for 2019 (G12) refers to the ST Loan in 2018
(E11) and to STRate (C4).
When you are done, choose Remove Arrows.
The auditing tools can also be used to see how each component of a formula evaluates,
one step at a time. For example, selecting the objective cell EndBalance (J21) and then
Evaluate Formula brings up the dialog box shown in Figure 4.11. Inside the evaluation box
is the formula in the selected cell, =J20 + SUM(C21:I21), with one component of the formula underlined. Pressing the Evaluate button evaluates the underlined component of the
formula (J20 becomes 10.273...) and then underlines the next component of the formula,
SUM(C21:I21). As shown at the bottom of the figure, each press of the Evaluate button
evaluates the underlined component of the formula and then underlines the next component
of the formula to be evaluated. Three presses fully evaluates the formula to show the end
result of 5.39. This tool can be particularly useful for evaluating complicated formulas to
make sure they are set up correctly.
4.4 Debugging a Spreadsheet Model 145
FIGURE 4.10
The spreadsheet obtained by using the Excel auditing tools to trace the precedents of the ST Interest (2019) calculation in cell G12 of
the spreadsheet in Figure 4.5.
A
1
B
C
D
E
F
G
H
I
J
K
L
Everglade Cash Flow Management Problem
2
3
LT Rate
5%
4
ST Rate
7%
(all cash figures in millions of dollars)
5
6
Start Balance
1
7
Minimum Cash
0.5
Cash
LT
ST
LT
ST
LT
ST
Ending
10
Year
Flow
Loan
Loan
Interest
Interest
Payback
Payback
Balance
11
2018
–8
4.65
2.85
8
9
–0.23
≥
0.5
–0.23
–0.37
–5.28
0.50
≥
0.5
–0.69
–9.88
0.50
≥
0.5
–0.23
–0.55
–7.81
0.50
≥
0.5
–0.23
–0.18
–2.59
0.50
≥
0.5
0
0
0.50
≥
0.5
–0.30
–4.23
2.74
≥
0.5
0.5
–2
5.28
2020
–4
9.88
14
2021
3
7.81
–0.23
15
2022
6
2.59
16
2023
3
0
17
2024
–4
4.23
–0.23
18
2025
7
0
–0.23
–2
0
2027
10
0
21
2028
0.5
0.50
2019
2026
≥
–2.85
13
20
Balance
0.50
–0.20
12
19
Minimum
–0.23
–0.23
0
0
0.51
≥
0
0
10.27
≥
0.5
–0.23
0
0
5.39
≥
0.5
–4.65
FIGURE 4.11
The dialog box for the
Evaluate Formula auditing
tool when applied to the
formula in EndBalance
(J21). Successive presses
of the Evaluate button
evaluate the underlined
portion of the formula,
step-by-step, as shown
below the dialog box.
Review
Questions
1. What is a good first step for debugging a spreadsheet model?
2. How do you toggle between viewing formulas and viewing values in output cells?
3. Which Excel tools can be used to trace the dependents or precedents for a given cell or
evaluate a formula in a cell?
146 Chapter Four The Art of Modeling with Spreadsheets
4.5 Summary
There is a considerable art to modeling well with spreadsheets. This chapter focuses on providing a
foundation for learning this art.
The general process of modeling in spreadsheets has four major steps: (1) plan the spreadsheet model,
(2) build the model, (3) test the model, and (4) analyze the model and its results. During the planning
step, it is helpful to begin by visualizing where you want to finish and then doing some calculations by
hand to clarify the needed computations before starting to sketch out a logical layout for the spreadsheet.
Then, when you are ready to undertake the building step, it is a good idea to start by building a small,
readily manageable version of the model before expanding the model to full-scale size. This enables you
to test the small version first to get all the logic straightened out correctly before expanding to a fullscale model and undertaking a final test. After completing all of this, you are ready for the analysis step,
which involves applying the model to evaluate proposed solutions and perhaps using Solver to optimize
the model.
Using this plan-build-test-analyze process should yield a spreadsheet model, but it doesn’t guarantee
that you will obtain a good one. Section 4.3 describes in detail the following guidelines for building
“good” spreadsheet models:
•
•
•
•
•
•
•
•
•
Enter the data first.
Organize and clearly identify the data.
Enter each piece of data into one cell only.
Separate data from formulas.
Keep it simple.
Use range names.
Use relative and absolute references to simplify copying formulas.
Use borders, shading, and colors to distinguish between cell types.
Show the entire model on the spreadsheet.
Even if all these guidelines are followed, a thorough debugging process may be needed to eliminate
the errors that lurk within the initial version of the model. It is important to check whether the output
cells are giving correct results for various values of the changing cells. Other items to check include
whether range names refer to the appropriate cells and whether formulas have been entered into output
cells correctly. Excel provides a number of useful features to aid in the debugging process. One is the
ability to toggle the worksheet between viewing the results in the output cells and the formulas entered
into those output cells. Several other helpful features are available with Excel’s auditing tools.
Glossary
auditing tools A set of tools provided by Excel
to aid in debugging a spreadsheet model.
(Section 4.4), 142
range name A descriptive name given to a range
of cells that immediately identifies what is there.
(Section 4.3), 139
toggle The act of switching back and forth
between viewing the results in the output cells
and viewing the formulas entered into those
­output cells. (Section 4.4), 142
Learning Aids for This Chapter
All learning aids are available at www.mhhe.com/Hillier6e.
Excel Files:
Excel Add-in:
Everglade Case Study
Analytic Solver
Wyndor Example
Everglade Problem 4.12
Everglade Problem 4.13
Solved Problems
The solutions are available at www.mhhe.com/Hillier6e.
4.S1. Production and Inventory Planning Model
Surfs Up produces high-end surfboards. A challenge faced
by Surfs Up is that their demand is highly seasonal. Demand
exceeds production capacity during the warm summer months,
but is very low in the winter months. To meet the high demand
during the summer, Surfs Up typically produces more surfboards
Chapter 4 Problems 147
than are needed in the winter months and then carries inventory
into the summer months. Their production facility can produce
at most 50 boards per month using regular labor at a cost of $125
each. Up to 10 additional boards can be produced by utilizing
overtime labor at a cost of $135 each. The boards are sold for
$200. Because of storage cost and the opportunity cost of capital, each board held in inventory from one month to the next
incurs a cost of $5 per board. Since demand is uncertain, Surfs
Up would like to maintain an ending inventory (safety stock) of
at least 10 boards during the warm months (May–September)
and at least 5 boards during the other months (October–April). It
is now the start of January and Surfs Up has 5 boards in inventory. The forecast of demand over the next 12 months is shown
in the following table. Formulate and solve a linear programming model in a spreadsheet to determine how many surfboards
should be produced each month to maximize total profit.
Forecasted Demand
Jan.
Feb.
Mar.
Apr.
May
June
July
Aug.
Sept.
Oct.
Nov.
Dec.
10
14
15
20
45
65
85
85
40
30
15
15
4.S2. Aggregate Planning: Manpower Hiring/
Firing/Training
Cool Power produces air-conditioning units for large commercial properties. Because of the low cost and efficiency of its
products, the company has been growing from year to year.
Also, seasonality in construction and weather conditions create production requirements that vary from month to month.
Cool Power currently has 10 fully trained employees working
in manufacturing. Each trained employee can work 160 hours
per month and is paid a monthly wage of $4,000. New trainees
can be hired at the beginning of any month. Because of their
lack of initial skills and required training, a new trainee provides only 100 hours of useful labor in the first month, but is
still paid a full monthly wage of $4,000. Furthermore, because
of required interviewing and training, there is a $2,500 hiring
cost for each employee hired. After one month, a trainee is considered fully trained. An employee can be fired at the beginning of any month, but must be paid two weeks of severance
pay ($2,000). Over the next 12 months, Cool Power forecasts
the labor requirements shown in the following table. Since
management anticipates higher requirements next year, Cool
Power would like to end the year with at least 12 fully trained
employees. How many trainees should be hired and/or workers
fired in each month to meet the labor requirements at the minimum possible cost? Formulate and solve a linear programming
spreadsheet model.
Labor Requirements (hours)
Jan.
Feb.
Mar.
Apr.
May
June
July
Aug.
Sept.
Oct.
Nov.
Dec.
1,600
2,000
2,000
2,000
2,800
3,200
3,600
3,200
1,600
1,200
800
800
Problems
An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.
4.1.
Consider the Everglade cash flow problem discussed
in this chapter. Suppose that extra cash is kept in an interest-­
bearing savings account. Assume that any cash left at the end of
a year earns 3 percent interest the following year. Make any necessary modifications to the spreadsheet and re-solve. (The original spreadsheet for this problem is available among the Excel
files for this chapter in www.mhhe.com/Hillier6e.)
4.2.*
The Pine Furniture Company makes fine country furniture. The company’s current product lines consist of end tables,
coffee tables, and dining room tables. The production of each of
these tables requires 8, 15, and 80 pounds of pine wood, respectively. The tables are handmade and require one hour, two hours,
and four hours, respectively. Each table sold generates $50, $100,
and $220 profit, respectively. The company has 3,000 pounds
of pine wood and 200 hours of labor available for the coming
week’s production. The chief operating officer (COO) has asked
you to do some spreadsheet modeling with these data to analyze
what the product mix should be for the coming week and make a
recommendation.
a. Visualize where you want to finish. What are the
decisions that need to be made? What should the
objective be? What numbers will the COO need?
b. Suppose that Pine Furniture were to produce three
end tables and three dining room tables. Calculate by
hand the amount of pine wood and labor that would
be required, as well as the profit generated from sales.
c. Make a rough sketch of a spreadsheet model, with
blocks laid out for the data cells, changing cells,
output cells, and objective cell.
d. Build a spreadsheet model and then solve it.
4.3.
Reboot, Inc., is a manufacturer of hiking boots. Demand
for boots is highly seasonal. In particular, the demand in the next
year is expected to be 3,000, 4,000, 8,000, and 7,000 pairs of
boots in quarters 1, 2, 3, and 4, respectively. With its current production facility, the company can produce at most 6,000 pairs of
boots in any quarter. Reboot would like to meet all the expected
demand, so it will need to carry inventory to meet demand in the
later quarters. Each pair of boots sold generates a profit of $20
per pair. Each pair of boots in inventory at the end of a quarter
incurs $8 in storage and capital recovery costs. Reboot has 1,000
pairs of boots in inventory at the start of quarter 1. Reboot’s top
management has given you the assignment of doing some spreadsheet modeling to analyze what the production schedule should
be for the next four quarters and making a recommendation.
a. Visualize where you want to finish. What are the
decisions that need to be made? What should the
objective be? What numbers are needed?
148 Chapter Four The Art of Modeling with Spreadsheets
b. Suppose that Reboot were to produce 5,000 pairs of
boots in each of the first two quarters. Calculate by
hand what the ending cash positions would be after
year 1 and year 2.
c. Make a rough sketch of a spreadsheet model, with
blocks laid out for the data cells, changing cells,
output cells, and objective cell.
d. Build a spreadsheet model for years 1 and 2, and
then thoroughly test the model.
e. Expand the model to full scale and then solve it.
4.4.*
The Fairwinds Development Corporation is considering taking part in one or more of three different development
­projects—A, B, and C—that are about to be launched. Each
project requires a significant investment over the next few years
and then would be sold upon completion. The projected cash
flows (in millions of dollars) associated with each project are
shown in the table below.
Year
1
2
3
4
5
6
Project A
Project B
Project C
−4
−6
−6
24
0
0
−8
−8
−4
−4
30
0
−10
−7
−7
−5
23
44
Fairwinds has $10 million available now and expects to receive
$6 million from other projects by the end of each year (1 through 6)
that would be available for the ongoing investments the following year in projects A, B, and C. By acting now, the company
may participate in each project either fully, fractionally (with other
development partners), or not at all. If Fairwinds participates at
less than 100 percent, then all the cash flows associated with that
project are reduced proportionally. Company policy requires ending each year with a cash balance of at least $1 million.
a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective
be? What numbers will top management need?
b. Suppose that Fairwinds were to participate in Project A fully and in Project C at 50 percent. Calculate
by hand the ending inventory, profit from sales, and
inventory costs for quarters 1 and 2.
c. Make a rough sketch of a spreadsheet model, with
blocks laid out for the data cells, changing cells,
output cells, and objective cell.
d. Build a spreadsheet model for quarters 1 and 2, and
then thoroughly test the model.
e. Expand the model to full scale, and then solve it.
4.5.
Read the referenced article that fully describes the management science study summarized in the application vignette
presented in Section 4.3. Briefly describe how spreadsheet modeling was applied in this study. Then list the various financial
and nonfinancial benefits that resulted from this study.
4.6.
Decorum, Inc., manufactures high-end ceiling fans. Their
sales are seasonal with higher demand in the warmer summer
months. Typically, sales average 400 units per month. However,
in the hot summer months (June, July, and August), sales spike up
to 600 units per month. Decorum can produce up to 500 units per
month at a cost of $300 each. By bringing in temporary workers,
Decorum can produce up to an additional 75 units at a cost of $350
each. Decorum sells the ceiling fans for $500 each. Decorum can
carry inventory from one month to the next, but at a cost of $20 per
ceiling fan per month. Decorum has 25 units in inventory at the start
of January. Assuming Decorum must produce enough ceiling fans to
meet demand, how many ceiling fans should Decorum produce each
month (using their regular labor force and/or temporary workers)
over the course of the next year so as to maximize their total profit?
a. Visualize where you want to finish. What are the
decisions that need to be made? What should the
objective be? What numbers will Decorum require?
b. Suppose Decorum builds 450 ceiling fans in January and 550 (utilizing temporary workers) in February. Calculate by hand the total costs for January
and February.
c. Make a rough sketch of a spreadsheet model, with
blocks laid out for the data cells, changing cells,
output cells, and objective cell.
d. Build a spreadsheet model for January and February
and thoroughly test the model.
e. Build and solve a linear programming spreadsheet
model to maximize the profit over all 12 months.
4.7.
Allen Furniture is a manufacturer of hand-crafted furniture. At the start of January, Allen employs 20 trained craftspeople.
They have forecasted their labor needs over the next 12 months
as shown in the following table. Each trained craftsperson provides 200 labor-hours per month and is paid a wage of $3,000 per
month. Hiring new craftspeople requires advertising, interviewing,
and then training at a cost of $2,500 per hire. New hires are called
apprentices for their first month. Apprentices spend their first
month observing and learning. They are paid $2,000 for the month,
but provide no labor. In their second month, apprentices are reclassified as trained craftspeople. The union contract allows for firing a
craftsperson at the beginning of a month, but $1,500 must be given
in severance pay. Moreover, at most 10% of the trained craftspeople
can be fired in any month. Allen would like to start next year with
at least 25 trained craftspeople. How many apprentices should be
hired and how many craftspeople should be fired in each month to
meet the labor requirements at the minimum possible cost?
a. Visualize where you want to finish. What are the
decisions that need to be made? What should the
objective be? What numbers will Allen require?
b. Suppose one apprentice is hired in January. Calculate by hand how many labor-hours would be available in January and February. Calculate by hand the
total costs in January and February.
c. Make a rough sketch of a spreadsheet model, with
blocks laid out for the data cells, changing cells,
output cells, and objective cell.
Labor Requirements (hours)
Jan.
Feb.
Mar.
Apr.
May
June
July
Aug.
Sept.
Oct.
Nov.
Dec.
3,400
4,000
4,200
4,200
3,000
2,800
3,000
4,000
4,500
5,000
5,200
4,800
Chapter 4 Problems 149
d. Build a spreadsheet model for January and February
and thoroughly test the model.
e. Build and solve a linear programming spreadsheet
model to maximize the profit over all 12 months.
4.8.
Refer to the scenario described in Problem 3.13
(­
Chapter 3), but ignore the instructions given there. Focus
instead on using spreadsheet modeling to address Web Mercantile’s problem by doing the following.
a. Visualize where you want to finish. What are the
decisions that need to be made? What should the
objective be? What numbers will Web Mercantile
require?
b. Suppose that Web Mercantile were to lease 30,000
square feet for all five months and then 20,000 additional square feet for the last three months. Calculate the total costs by hand.
c. Make a rough sketch of a spreadsheet model, with
blocks laid out for the data cells, changing cells,
output cells, and objective cell.
d. Build a spreadsheet model for months 1 and 2, and
then thoroughly test the model.
e. Expand the model to full scale, and then solve it.
4.9.*
Refer to the scenario described in Problem 3.15
(­
Chapter 3), but ignore the instructions given there. Focus
instead on using spreadsheet modeling to address Larry Edison’s
problem by doing the following.
a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective
be? What are the numbers that Larry will require?
b. Suppose that Larry were to hire three full-time
workers for the morning shift, two for the afternoon
shift, and four for the evening shift, as well as three
A
1
part-time workers for each of the four shifts. Calculate by hand how many workers would be working
at each time of the day and what the total cost would
be for the entire day.
c. Make a rough sketch of a spreadsheet model, with
blocks laid out for the data cells, changing cells,
output cells, and objective cell.
d. Build a spreadsheet model and then solve it.
4.10.
Refer to the scenario described in Problem 3.19
(­Chapter 3), but ignore the instructions given there. Focus
instead on using spreadsheet modeling to address Al Ferris’s
problem by doing the following.
a. Visualize where you want to finish. What are the
decisions that need to be made? What should the
objective be? What numbers will Al require?
b. Suppose that Al were to invest $20,000 each in
investment A (year 1), investment B (year 2), and
investment C (year 2). Calculate by hand what the
ending cash position would be after each year.
c. Make a rough sketch of a spreadsheet model, with
blocks laid out for the data cells, changing cells,
output cells, and objective cell.
d. Build a spreadsheet model for years 1 through 3,
and then thoroughly test the model.
e. Expand the model to full scale, and then solve it.
4.11.
In contrast to the spreadsheet model for the Wyndor
Glass Co. product-mix problem shown in Figure 4.6, the following spreadsheet is an example of a poorly formulated spreadsheet
model for this same problem. Referring to Section 4.3, identify
the guidelines violated by the model below. Then, explain how
each guideline has been violated and why the model in Figure
4.6 is a better alternative.
B
C
D
Wyndor Glass Co. (Poor Formulation)
2
3
Doors Produced
2
4
Windows Produced
6
5
Hours Used (Plant 1)
2
6
Hours Used (Plant 2)
12
7
Hours Used (Plant 3)
18
8
Total Profit
Solver Parameters
Set Objective Cell: C8
To: Max
By Changing Variable Cells:
C3:C4
Subject to the Constraints:
C5 <= 4
C6 <= 12
C7 <= 18
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
$3,600
B
C
5
Hours Used (Plant 1)
=1*C3+0*C4
6
Hours Used (Plant 2)
=0*C3+2*C4
7
8
Hours Used (Plant 3)
=3*C3+2*C4
Total Profit
=300*C3+500*C4
150 Chapter Four The Art of Modeling with Spreadsheets
4.12.
Refer to the spreadsheet file named “Everglade Problem
4.12,” available in www.mhhe.com/Hillier6e. This file contains a
formulation of the Everglade problem considered in this chapter.
However, three errors are included in this formulation. Use the ideas
presented in Section 4.4 for debugging a spreadsheet model to find
the errors. In particular, try different trial values for which you can
predict the correct results, use the toggle to examine all the formulas, and use the auditing tools to check precedence and dependence
relationships among the various changing cells, data cells, and output cells. Describe the errors found and how you found them.
4.13.
Refer to the spreadsheet file named “Everglade Problem
4.13,” available in www.mhhe.com/Hillier6e. This file contains a
formulation of the Everglade problem considered in this chapter.
However, three errors are included in this formulation. Use the ideas
presented in Section 4.4 for debugging a spreadsheet model to find
the errors. In particular, try different trial values for which you can
predict the correct results, use the toggle to examine all the formulas, and use the auditing tools to check precedence and dependence
relationships among the various changing cells, data cells, and output cells. Describe the errors found and how you found them.
Case 4-1
Prudent Provisions for Pensions
Among its many financial products, the Prudent Financial Services Corporation (normally referred to as PFS) manages a wellregarded pension fund that is used by a number of companies to
provide pensions for their employees. PFS’s management takes
pride in the rigorous professional standards used in operating
the fund. Since the near-collapse of the financial markets during
the protracted Great Recession that began in late 2007, PFS has
redoubled its efforts to provide prudent management of the fund.
It is now December 2017. The total pension payments that
will need to be made by the fund over the next 10 years are
shown in the following table.
Year
Pension Payments ($ millions)
2018
2019
2020
2021
2022
2023
2024
2025
2026
2027
8
12
13
14
16
17
20
21
22
24
By using interest as well, PFS currently has enough liquid
assets to meet all these pension payments. Therefore, to safeguard
the pension fund, PFS would like to make a number of investments
whose payouts would match the pension payments over the next
10 years. The only investments that PFS trusts for the pension fund
are a money market fund and bonds. The money market fund pays
an annual interest rate of 2 percent. The characteristics of each unit
of the four bonds under consideration are shown in the table below.
Bond 1
Bond 2
Bond 3
Bond 4
Current
Price
Coupon
Rate
Maturity
Date
Face
Value
$980
920
750
800
4%
2
0
3
Jan. 1, 2019
Jan. 1, 2021
Jan. 1, 2023
Jan. 1, 2026
$1,000
1,000
1,000
1,000
All of these bonds will be available for purchase on January 1,
2018, in as many units as desired. The coupon rate is the percentage of the face value that will be paid in interest on January 1 of
each year, starting one year after purchase and continuing until
(and including) the maturity date. Thus, these interest payments
on January 1 of each year are in time to be used toward the pension payments for that year. Any excess interest payments will
be deposited into the money market fund. To be conservative
in its financial planning, PFS assumes that all the pension payments for the year occur at the beginning of the year immediately
after these interest payments (including a year’s interest from the
money market fund) are received. The entire face value of a bond
also will be received on its maturity date. Since the current price
of each bond is less than its face value, the actual yield of the
bond exceeds its coupon rate. Bond 3 is a zero-­coupon bond, so
it pays no interest but instead pays a face value on the maturity
date that greatly exceeds the purchase price.
PFS would like to make the smallest possible investment
(including any deposit into the money market fund) on January
1, 2018, to cover all its required pension payments through 2027.
Some spreadsheet modeling needs to be done to see how to do this.
a. Visualize where you want to finish. What are the decisions
that need to be made? What should the objective be? What
numbers are needed by PFS management?
b. Suppose that PFS were to invest $30 million in the money
market fund and purchase 10,000 units each of bond 1 and
bond 2 on January 1, 2018. Calculate by hand the payments
received from bonds 1 and 2 on January 1 of 2019 and 2020.
Also calculate the resulting balance in the money market
fund on January 1 of 2018, 2019, and 2020 after receiving
these payments, making the pension payments for the year,
and depositing any excess into the money market fund.
c. Make a rough sketch of a spreadsheet model, with blocks laid out
for the data cells, changing cells, output cells, and objective cell.
d. Build a spreadsheet model for years 2018 through 2020, and
then thoroughly test the model.
e. Expand the model to consider all years through 2027, and
then solve it.
Additional Case
Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website,
cases.ivey.uwo.ca/cases, in the segment of the CaseMate area
designated for this book.
Chapter Five
What-If Analysis for
Linear Programming
Learning Objectives
After completing this chapter, you should be able to
1. Explain what is meant by what-if analysis.
2. Summarize the benefits of what-if analysis.
3. Enumerate the different kinds of changes in the model that can be considered by what-if
analysis.
4. Describe how the spreadsheet formulation of the problem can be used to perform any of
these kinds of what-if analysis.
5. Use parameter analysis reports generated by Analytic Solver to systematically investigate
the effect of changing either one or two data cells to various other trial values.
6. Find how much any single coefficient in the objective function can change without
changing the optimal solution.
7. Evaluate simultaneous changes in objective function coefficients to determine whether
the changes are small enough that the original optimal solution must still be optimal.
8. Predict how the value in the objective cell would change if a small change were to be
made in the right-hand side of one or more of the functional constraints.
9. Find how much the right-hand side of a single functional constraint can change before
this prediction becomes no longer valid.
10. Evaluate simultaneous changes in right-hand sides to determine whether the changes
are small enough that this prediction must still be valid.
11. Describe the goal of robust optimization and how it is implemented with independent
parameters.
12. Use chance constraints to deal with constraints that actually can be violated a little bit.
Chapters 2 to 4 have described and illustrated how to formulate a linear programming model
on a spreadsheet to represent a variety of managerial problems, and then how to use Solver to
find an optimal solution for this model. You might think that this would finish our story about
linear programming: Once the manager learns the optimal solution, she would immediately
implement this solution and then turn her attention to other matters. However, this is not the
case. The enlightened manager demands much more from linear programming, and linear
programming has much more to offer her—as you will discover in this chapter.
An optimal solution is only optimal with respect to a particular mathematical model that
provides only a rough representation of the real problem. A manager is interested in much
more than just finding such a solution. The purpose of a linear programming study is to
help guide management’s final decision by providing insights into the likely consequences of
pursuing various managerial options under a variety of assumptions about future conditions.
Most of the important insights are gained while conducting analysis after finding an optimal
solution for the original version of the basic model. This analysis is commonly referred to as
what-if analysis because it involves addressing some questions about what would happen to
151
152 Chapter Five What-If Analysis for Linear Programming
the optimal solution if different assumptions were made about future conditions. Spreadsheets
play a central role in addressing these what-if questions.
This chapter focuses on the types of information provided by what-if analysis and why
it is valuable to managers. The first section provides an overview. Section 5.2 returns to the
Wyndor Glass Co. product-mix case study (Section 2.1) to describe the what-if analysis that
is needed in this situation. The next four sections then perform this what-if analysis, using
a variety of procedures that are applicable to any linear programming problem. Section 5.7
extends this approach further to provide a way of obtaining a solution that is virtually guaranteed to be feasible. Section 5.8 describes how to deal with constraints that actually can be
violated a little bit without serious complications.
5.1
THE IMPORTANCE OF WHAT-IF ANALYSIS TO MANAGERS
In real applications, many
of the numbers in the
model may be only rough
estimates.
What happens to the optimal solution if an error
is made in estimating a
parameter of the model?
The examples and problems in the preceding chapters on linear programming have provided
the data needed to determine precisely all the numbers that should go into the data cells for
the spreadsheet formulation of the linear programming model. (Recall that these numbers are
referred to as the parameters of the model.) Real applications seldom are this straightforward. Substantial time and effort often are needed to track down the needed data. Even then, it
may be possible to develop only rough estimates of the parameters of the model.
For example, in the Wyndor case study, two key parameters of the model are the coefficients in the objective function that represent the unit profits of the two new products. These
parameters were estimated to be $300 for the doors and $500 for the windows. However,
these unit profits depend on many factors—the costs of raw materials, production, shipping,
advertising, and so on, as well as such things as the market reception to the new products and
the amount of competition encountered. Some of these factors cannot be estimated with real
accuracy until long after the linear programming study has been completed and the new products have been on the market for some time.
Therefore, before Wyndor’s management makes a decision on the product mix, it will want
to know what the effect would be if the unit profits turn out to differ significantly from the
estimates. For example, would the optimal solution change if the unit profit for the doors
turned out to be $200 instead of the estimate of $300? How inaccurate can the estimate be in
either direction before the optimal solution changes?
Such questions are addressed in Section 5.3 when only one estimate is inaccurate.
­Section 5.4 will address similar questions when multiple estimates are inaccurate.
If the optimal solution will remain the same over a wide range of values for a particular
coefficient in the objective function, then management will be content with a fairly rough
estimate for this coefficient. On the other hand, if even a small error in the estimate would
change the optimal solution, then management will want to take special care to refine this
estimate. Management sometimes will get involved directly in adjusting such estimates to its
satisfaction.
Here then is a summary of the first benefit of what-if analysis:
1. Typically, many of the parameters of a linear programming model are only estimates of quantities (e.g., unit profits) that cannot be determined precisely at this time. What-if analysis
reveals how close each of these estimates needs to be to avoid obtaining an erroneous optimal
solution, and therefore pinpoints the sensitive parameters (those parameters where extra
care is needed to refine their estimates because even small changes in their values can change
the optimal solution).
Several sections describe how what-if analysis provides this benefit for the most important
parameters. Sections 5.3 and 5.4 do this for the coefficients in the objective function (these
numbers typically appear in the spreadsheet in the row for the unit contribution of each activity
toward the overall measure of performance). Sections 5.5 and 5.6 do the same for the right-hand
sides of the functional constraints (these are the numbers that typically are in the right-hand
column of the spreadsheet just to the right of the ≤, ≥, or = signs).
Businesses operate in a dynamic environment. Even when management is satisfied with the
current estimates and implements the corresponding optimal solution, conditions may change
5.1 The Importance of What-If Analysis to Managers 153
What happens to the optimal solution if conditions
change in the future?
What happens if managerial
policy decisions change?
later. For example, suppose that Wyndor’s management is satisfied with $300 as the estimate
of the unit profit for the doors, but increased competition later forces a price reduction that
reduces this unit profit. Does this change the optimal product mix? The what-if analysis shown
in Section 5.3 immediately indicates in advance which new unit profits would leave the optimal product mix unchanged, which can help guide management in its new pricing decision.
Furthermore, if the optimal product mix is unchanged, then there is no need to solve the model
again with the new coefficient. Avoiding solving the model again is no big deal for the tiny twovariable Wyndor problem, but it is extremely welcome for real applications that may have hundreds or thousands of constraints and variables. In fact, for such large models, it may not even
be practical to re-solve the model repeatedly to consider the many possible changes of interest.
Thus, here is the second benefit of what-if analysis:
2. If conditions change after the study has been completed (a common occurrence), what-if
analysis leaves signposts that indicate (without solving the model again) whether a resulting
change in a parameter of the model changes the optimal solution.
Again, several subsequent sections describe how what-if analysis does this.
These sections focus on studying how changes in the parameters of a linear programming
model affect the optimal solution. This type of what-if analysis commonly is referred to as
sensitivity analysis, because it involves checking how sensitive the optimal solution is to the
value of each parameter. Sensitivity analysis is a vital part of what-if analysis.
However, rather than being content with the passive sensitivity analysis approach of checking the effect of parameter estimates being inaccurate, what-if analysis often goes further to
take a proactive approach. An analysis may be made of various possible managerial actions
that would result in changes to the model.
A prime example of this proactive approach arises when certain parameters of the model
represent managerial policy decisions rather than quantities that are largely outside the control of management. For example, for the Wyndor product-mix problem, the right-hand sides
of the three functional constraints (4, 12, 18) represent the number of hours of production
time in the three respective plants being made available per week for the production of the
two new products. Management can change these three resource amounts by altering the
production levels for the old products in these plants. Therefore, after learning the optimal
solution, management will want to know the impact on the profit from the new products if
these resource amounts are changed in certain ways. One key question is how much this
profit can be increased by increasing the available production time for the new products in
just one of the plants. Another is how much this profit can be increased by simultaneously
making helpful changes in the available production times in all the plants. If the profit from
the new products can be increased enough to more than compensate for the profit lost by
decreasing the production levels for certain old products, management probably will want to
make the change.
We now can summarize the third benefit of what-if analysis:
3. When certain parameters of the model represent managerial policy decisions, what-if analysis provides valuable guidance to management regarding the impact of altering these policy
decisions.
Sections 5.5 and 5.6 will explore this benefit further.
What-if analysis sometimes goes even further in providing helpful guidance to m
­ anagement.
For example, when there is considerable uncertainty about what the actual values of the model
parameters will turn out to be, management might want to identify a solution that is virtually guaranteed to be both feasible and nearly optimal for all plausible combinations of the
actual values of these parameters. Section 5.7 will introduce a robust optimization technique
for doing this. Section 5.8 then will describe how to use chance constraints to deal with constraints that can be violated a little bit without serious complications.
Throughout this chapter, the focus is on how and when to use what-if analysis while applying linear programming. What-if analysis also is important to managers when applying the
various other techniques of management science. Although the procedures may need to be
adapted to fit other kinds of applications, this chapter illustrates the importance of including
what-if analysis in almost any management science study.
154 Chapter Five What-If Analysis for Linear Programming
Review
Questions
1.
2.
3.
4.
5.
6.
7.
8.
5.2
What are the parameters of a linear programming model?
How can inaccuracies arise in the parameters of a model?
What does what-if analysis reveal about the parameters of a model that are only estimates?
Is it always inappropriate to make only a fairly rough estimate for a parameter of a model?
Why?
How is it possible for the parameters of a model to be accurate initially and then become inaccurate at a later date?
How does what-if analysis help management prepare for changing conditions?
What is meant by sensitivity analysis?
For what kinds of managerial policy decisions does what-if analysis provide guidance?
CONTINUING THE WYNDOR CASE STUDY
We now return to the case study introduced in Section 2.1 involving the Wyndor Glass Co.
product-mix problem.
To review briefly, recall that the company is preparing to introduce two new products:
• An 8-foot glass door with aluminum framing.
• A 4-foot × 6-foot double-hung wood-framed window.
To analyze which mix of the two products would be most profitable, the company’s Management Science Group introduced two decision variables:
D = Production rate of this new kind of door
W = Production rate of this new kind of window
where this rate measures the number of units produced per week. Three plants will be involved
in the production of these products. Based on managerial decisions regarding how much these
plants will continue to be used to produce current products, the number of hours of production
time per week being made available in Plants 1, 2, and 3 for the new products is 4, 12, and 18,
respectively. After obtaining rough estimates that the profit per unit will be $300 for the doors
and $500 for the windows, the Management Science Group then formulated the linear programming model shown in Figure 2.12 and repeated here in Figure 5.1, where the objective is to
choose the values of D and W in the changing cells UnitsProduced (C12:D12) so as to maximize
the total profit (per week) given in the objective cell TotalProfit (G12). Applying Solver to this
model yielded the optimal solution shown on this spreadsheet and summarized as follows.
Optimal Solution
D =2
(​​ ​​Produce 2 doors per week.​)​​​
​​ W​ ​  =​ ​  6​ ​
​  (​​ ​​Produce 6 windows per week.​
​ )​​​​
​​
Profit = 3,600 (​​ ​​The estimated total weekly profit is $3,600.​)​​​
However, this optimal solution assumes that all the estimates that provide the parameters of
the model (as shown in the UnitProfit (C4:D4), HoursUsedPerUnitProduced (C7:D9), and
HoursAvailable (G7:G9) data cells) are accurate.
The head of the Management Science Group, Lisa Taylor, now is ready to meet with management to discuss the group’s recommendation that the above product mix be used.
Management’s Discussion of the Recommended Product Mix
Lisa Taylor (head of Management Science Group): I asked for this meeting so we could
explore what questions the two of you would like us to pursue further. In particular, I am
especially concerned that we weren’t able to better pin down just what the numbers should
be to go into our model. Which estimates do you think are the shakiest?
Bill Tasto (vice president for manufacturing): Without question, the estimates of the
unit profits for the two products. Since the products haven’t gone into production yet, all
we could do is analyze the data from similar current products and then try to project what
5.2 Continuing the Wyndor Case Study 155
FIGURE 5.1
The spreadsheet model
and its optimal solution
for the original Wyndor
problem before beginning
what-if analysis.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
2
≤
4
8
Plant 2
0
2
12
≤
12
9
Plant 3
3
2
18
≤
18
Doors
Windows
Total Profit
2
6
$3,600
10
11
12
Units Produced
Solver Parameters
Set Objective Cell: TotalProfit
To: Max
By Changing Variable Cells:
UnitsProduced
Subject to the Constraints:
HoursUsed <= HoursAvailable
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
E
5
6
7
8
9
Used
=SUMPRODUCT(C7:D7, UnitsProduced)
=SUMPRODUCT(C8:D8, UnitsProduced)
=SUMPRODUCT(C9:D9, UnitsProduced)
G
11
12
The allowable range for a
unit profit indicates how far
its estimate can be off without affecting the optimal
product mix.
Hours
Total Profit
=SUMPRODUCT(UnitProfit, UnitsProduced)
Range Name
Cells
DoorsProduced
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
TotalProfit
UnitProfit
UnitsProduced
WindowsProduced
C12
G7:G9
E7:E9
C7:D9
G12
C4:D4
C12:D12
D12
the changes would be for these new products. We have some numbers, but they are pretty
rough. We would need to do a lot more work to pin down the numbers better.
John Hill (president): We may need to do that. Lisa, do you have a way of checking how
far off one of these estimates can be without changing the optimal product mix?
Lisa: Yes, we do. We can quickly find what we call the allowable range for each unit profit.
As long as the true value of the unit profit is within this allowable range, and the other unit
profit is correct, the optimal product mix will not change. If this range is pretty wide, you
don’t need to worry about refining the estimate of the unit profit. However, if the range is
quite narrow, then it is important to pin down the estimate more closely.
John: What happens if both estimates are off?
Lisa: We can provide a way of checking whether the optimal product mix might change for
any new combination of unit profits you think might be the true one.
John: Great. That’s what we need. There’s also another thing. Bill gave you the numbers for
how many hours of production time we’re making available per week in the three plants for
these new products. I noticed you used these numbers on your spreadsheet.
Lisa: Yes. They’re the right-hand sides of our constraints. Is something wrong with these
numbers?
156 Chapter Five What-If Analysis for Linear Programming
John: No, but we would like your group to provide us with some analysis of what the effect
would be if we change any one of those numbers. How much more profit could we get from
the new products for each additional hour of production time per week we provide in one of
the plants? That sort of thing.
Lisa: Yes, we can get that analysis to you right away also.
John: We might also be interested in changing the available production hours for two or
three of the plants. Other key numbers that can only be estimated are the production rates of
the doors and windows in plant 3. What happens if those estimates are uncertain?
Lisa: No problem. We’ll give you information about all of that as well.
Summary of Management’s What-If Questions
Here is a summary of John Hill’s what-if questions that Lisa and her group will be addressing
in the coming sections.
1. What happens if the estimate of the unit profit of one of Wyndor’s new products is inaccurate? (Section 5.3)
2. What happens if the estimates of the unit profits of both of Wyndor’s new products are
inaccurate? (Section 5.4)
3. What happens if a change is made in the number of hours of production time per week
being made available to Wyndor’s new products in one of the plants? (Section 5.5)
4. What happens if simultaneous changes are made in the number of hours of production time
per week being made available to Wyndor’s new products in all the plants? (Section 5.6)
5. What happens if the production rates of doors and windows at plant 3 are uncertain?
(­Sections 5.7 and 5.8)
Review
Questions
1. Which estimates of the parameters in the linear programming model for the Wyndor problem
are most questionable?
2. Which numbers in this model represent tentative managerial decisions that management
might want to change after receiving the Management Science Group’s analysis?
5.3 THE EFFECT OF CHANGES IN ONE OBJECTIVE
FUNCTION COEFFICIENT
Section 5.1 began by discussing the fact that many of the parameters of a linear programming
model typically are only estimates of quantities that cannot be determined precisely at the
time. What-if analysis (or sensitivity analysis in particular) reveals how close each of these
estimates needs to be to avoid obtaining an erroneous optimal solution.
We focus in this section on how sensitivity analysis does this when the parameters involved are
coefficients in the objective function. (Recall that each of these coefficients gives the unit contribution of one of the activities toward the overall measure of performance.) In the process, we will
address the first of the what-if questions posed by Wyndor management in the preceding section.
Question 1: What happens if the estimate of the unit profit of one of Wyndor’s new products is
inaccurate?
To start this process, first consider the question of what happens if the estimate of $300
for the unit profit for Wyndor’s new kind of door is inaccurate. To address this question, let
​ ​  D​​ = Unit profit for the new kind of door
P
​ ​​
​  ​
​
​​
​ = Cell C4 in the spreadsheet (see Figure 5.1)
Although PD = $300 in the current version of Wyndor’s linear programming model, we now
want to explore how much larger or how much smaller PD can be and still have (D, W) = (2, 6)
as the optimal solution. In other words, how much can the estimate of $300 for the unit profit
for these doors be off before the model will give an erroneous optimal solution?
Using the Spreadsheet to Do Sensitivity Analysis
One of the great strengths of a spreadsheet is the ease with which it can be used interactively to
perform various kinds of what-if analysis, including the sensitivity analysis being considered
5.3 The Effect of Changes in One Objective Function Coefficient 157
Run Solver again and the
spreadsheet immediately
reveals the effect of changing any values in the data
cells.
in this section. Once Solver has been set up to obtain an optimal solution, you can immediately
find out what would happen if one of the parameters of the model were to be changed to some
other value. All you have to do is make this change on the spreadsheet and then run Solver again.
To illustrate, Figure 5.2 shows what would happen if the unit profit for doors were to be
decreased from PD = $300 to PD = $200. Comparing with Figure 5.1, there is no change at
all in the optimal solution. In fact, the only changes in the new spreadsheet are the new value
of PD in cell C4 and a decrease of $200 in the total profit shown in cell G12 (because each
of the two doors produced per week provides $100 less profit). Because the optimal solution
does not change, we now know that the original estimate of PD = $300 can be considerably
too high without invalidating the model’s optimal solution.
But what happens if this estimate is too low instead? Figure 5.3 shows what would happen
if PD were to be increased to PD = $500. Again, there is no change in the optimal solution.
Because the original value of PD = $300 can be changed considerably in either direction
without changing the optimal solution, PD is said to be not a sensitive parameter. It is not
necessary to pin down this estimate with great accuracy to have confidence that the model is
providing the correct optimal solution.
This may be all the information that is needed about PD. However, if there is a good
­possibility that the true value of PD will turn out to be outside this broad range from $200 to
$500, further investigation would be desirable. How much higher or lower can PD be before
the optimal solution would change?
FIGURE 5.2
The revised Wyndor problem where the estimate of
the unit profit for doors
has been decreased from
PD = $300 to PD = $200,
but no change occurs in
the optimal solution.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
Unit Profit
4
Doors
Windows
$200
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
2
≤
4
8
Plant 2
0
2
12
≤
12
9
Plant 3
3
2
18
≤
18
Doors
Windows
Total Profit
2
6
$3,400
C
D
10
11
12
FIGURE 5.3
The revised Wyndor problem where the estimate of
the unit profit for doors
has been increased from
PD = $300 to PD = $500,
but no change occurs in
the optimal solution.
Units Produced
A
1
B
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$500
$500
5
6
7
Hours Used per Unit Produced
Plant 1
1
0
Hours
Hours
Used
Available
2
≤
4
12
18
8
Plant 2
0
2
12
≤
9
Plant 3
3
2
18
≤
Doors
Windows
Total Profit
2
6
$4,000
10
11
12
Units Produced
158 Chapter Five What-If Analysis for Linear Programming
Figure 5.4 demonstrates that the optimal solution would indeed change if PD were increased
all the way up to PD = $1,000. Thus, we now know that this change occurs somewhere between
$500 and $1,000 during the process of increasing PD.
Using a Parameter Analysis Report Generated by Analytic
Solver to Do Sensitivity Analysis Systematically
To pin down just when the optimal solution will change, we could continue selecting new values
of PD at random. However, a better approach is to systematically consider a range of values of PD.
The Analytic Solver, first introduced in Section 2.6, can generate a parameter analysis
report that is designed to do just this sort of analysis. Instructions for downloading this software are available in the preface and at www.mhhe.com/Hillier6e.
In preparation for running a parameter analysis report, several additional range names were
defined for specific cells that will be changed (e.g., UnitProfitPerDoor for C4) and for the specific changing cells (DoorsProduced and WindowsProduced for C12 and D12, respectively).
This allows the resulting report to show the results with more informative names for each of
these cells.
The data cell containing a parameter that will be systematically varied (UnitProfitPerDoor, or
C4, in this case) is referred to as a parameter cell. A parameter analysis report is used to show
the results in the changing cells and/or the objective cell for various trial values in the parameter
cell. For each trial value, these results are obtained by using Solver to re-solve the problem.
To generate a parameter analysis report, the first step is to define the parameter cell. In
this case, select cell C4 (UnitProfitPerDoor) and choose Parameters > Optimization on the
Analytic Solver ribbon. In the parameter cell dialog box, shown in Figure 5.5, enter the range
FIGURE 5.4
The revised Wyndor
problem where the estimate of the unit profit for
doors has been increased
from PD = $300 to PD =
$1,000, which results in
a change in the optimal
solution.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$1,000
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
4
≤
4
8
Plant 2
0
2
6
12
9
Plant 3
3
2
18
≤
≤
Doors
Windows
Total Profit
4
3
$5,500
18
10
11
12
FIGURE 5.5
The parameter cell dialog
box for PD specifies here
that this parameter cell for
the Wyndor problem will
be systematically varied
from $100 to $1,000.
Units Produced
5.3 The Effect of Changes in One Objective Function Coefficient 159
of trial values for the parameter cell. The entries shown specify that PD will be systematically
varied from $100 to $1,000. If desired, additional parameter cells could be defined in this
same way, but we will not do so at this point.
Next choose Reports > Optimization > Parameter Analysis on the Analytic Solver ribbon. This brings up the dialog box shown in Figure 5.6 that allows you to specify which
parameter cells to vary and which results to show. The choice of which parameter cells to
vary is made under Parameters in the bottom half of the dialog box. Clicking on (>>) will
select all of the parameter cells defined so far (moving them to the box on the right). In the
­Wyndor ­example, only one parameter has been defined, so this causes the single parameter cell
­(UnitProfitPerDoor) to appear on the right. If more parameter cells had been defined, particular parameter cells can be chosen for immediate analysis by clicking on the + next to Wyndor
to reveal the list of parameter cells that have been defined in the Wyndor spreadsheet. Clicking
on (>) then moves individual parameter cells to the list on the right.
The choice of which results to show as the parameter cell is varied is made in the upper
half of the dialog box. Clicking on (>>) will cause all of the changing cells (DoorsProduced
and WindowsProduced) and the objective cell (Total Profit) to appear in the list on the right.
To instead choose a subset of these cells, click on the small + next to Variables (or Objective)
to reveal a list of all the changing cells (or objective cell) and then click on > to move that
changing cell (or objective cell) to the right.
Finally, enter the number of Major Axis Points to specify how many different values of
the parameter cell will be shown in the parameter analysis report. The values will be spread
evenly between the lower and upper values specified in the parameters cell dialog box in
­Figure 5.5. With 10 major axis points, a lower value of $100, and an upper value of $1000, the
parameter analysis report will show results for PD of $100, $200, $300, . . ., $1000.
FIGURE 5.6
The dialog box for the
parameter analysis report
specifies here for the
Wyndor problem that
the UnitProfitPerDoor
parameter cell will be
varied and that results
from all the changing
cells (DoorsProduced
and WindowsProduced)
and the objective cell
(Total-Profit) will be
shown.
160 Chapter Five What-If Analysis for Linear Programming
The allowable range for a
coefficient in the objective
function is the range of values for this coefficient over
which the optimal solution for the original model
remains optimal.
Clicking on the OK button generates the parameter analysis report shown in Figure 5.7. One
at a time, the trial values listed in the first column of the table are put into the parameter cell
(UnitProfitPerDoor) and then Solver is called on to re-solve the problem. The optimal results
for that particular trial value of the parameter cell are then shown in the remaining columns—
DoorsProduced, WindowsProduced, and TotalProfit. This is repeated automatically for each
remaining trial value of the parameter cell. The end result (which happens very quickly for
small problems) is the completely-filled-in parameter analysis report shown in Figure 5.7.
The parameter analysis report reveals that the optimal solution remains the same all the
way from PD = $100 (and perhaps lower) to PD = $700, but that a change occurs somewhere
between $700 and $800. We next could systematically consider values of PD between $700
and $800 to determine more closely where the optimal solution changes. However, here is a
shortcut. The range of values of PD over which (D, W) = (2, 6) remains as the optimal solution is referred to as the allowable range for an objective function coefficient, or just
the allowable range for short. Upon request, Solver will provide a report called the sensitivity
report that reveals exactly what this allowable range is.
Using the Sensitivity Report to Find the Allowable Range
As was shown in Figure 2.12, when Excel’s Solver gives the message that it has found a solution,
it also gives on the right a list of three reports that can be provided. By selecting the ­second one
(labeled Sensitivity), you will obtain the sensitivity report. With Analytic Solver, this same report
is obtained by choosing Reports > Optimization > Sensitivity from the A
­ nalytic Solver ribbon.
Figure 5.8 shows the relevant part of this report for the Wyndor problem. The Final Value
column indicates the optimal solution. The next column gives the reduced costs, which can provide some useful information when any of the changing cells equal zero in the optimal solution,
which is not the case here. (For a zero-valued changing cell, the corresponding reduced cost can
FIGURE 5.7
The parameter analysis report that shows the effect of systematically varying the estimate of the unit profit for doors for the
Wyndor problem.
A
B
C
D
1
UnitProfitPerDoor
DoorsProduced
WindowsProduced
TotalProfit
2
$100
2
6
$3,200
3
$200
2
6
$3,400
4
$300
2
6
$3,600
5
$400
2
6
$3,800
6
$500
2
6
$4,000
$4,200
7
$600
2
6
8
$700
2
6
$4,400
9
$800
4
3
$4,700
10
$900
4
3
$5,100
11
$1,000
4
3
$5,500
FIGURE 5.8
Part of the sensitivity
report generated by Solver
for the original Wyndor
problem (Figure 5.1),
where the last three columns enable identifying
the allowable ranges for
the unit profits for doors
and windows.
Variable Cells
Cell
Name
Final
Value
Reduced
Cost
Objective
Coefficient
Allowable
Increase
Allowable
Decrease
$C$12
$D$12
DoorsProduced
WindowsProduced
2
6
0
0
300
500
450
IE + 30
300
300
5.3 The Effect of Changes in One Objective Function Coefficient 161
Reduced costs are described
in the supplement to this
chapter at www.mhhe.
com/Hillier6e. Don’t
worry about this relatively
technical subject (unless
your instructor assigns this
supplement).
The sensitivity report
generated by Solver reveals
the allowable range for each
coefficient in the objective
function.
be used to determine what the effect would be of either increasing that changing cell or making
a change in its coefficient in the objective function. Because of the relatively technical nature of
these interpretations of reduced costs, we will not discuss them further here, but will provide a
full explanation in the supplement to this chapter at www.mhhe.com/­Hillier6e. The next three
columns provide the information needed to identify the allowable range for each coefficient in
the objective function. The Objective Coefficient column gives the current value of each coefficient, and then the next two columns give the allowable increase and the allowable decrease
from this value to remain within the allowable range.
For example, consider PD, the coefficient of D in the objective function. Since D is the
production rate for these special doors, the Doors row in the table provides the following
information (without the dollar sign) about PD:
Current value of ​P​  D​​:
300
​
Allowable increase in ​P​  D​​: 450
So ​P​  D​​  ≤ 300 + 450 = 750
​​
​  ​  ​  ​
​
​  ​
​ ​​​
Allowable decrease in ​P​  D​​: 300
So ​P​  D​​ ≥ 300 − 300 = 0
Allowable range for ​P​  D​​: 0 ≤ ​P​  D​​ ≤ 750 ​
Therefore, if PD is changed from its current value (without making any other change in the
model), the current solution (D, W) = (2, 6) will remain optimal so long as the new value of
PD is within this allowable range.
Figure 5.9 provides graphical insight into this allowable range. For the original value of
PD = 300, the solid line in the figure shows the slope of the objective function line passing
through (2, 6). At the lower end of the allowable range, when PD = 0, the objective function
line that passes through (2, 6) now is line B in the figure, so every point on the line segment
between (0, 6) and (2, 6) is an optimal solution. For any value of PD < 0, the objective function line will have rotated even further so that (0, 6) becomes the only optimal solution. At
the upper end of the allowable range, when PD = 750, the objective function line that passes
through (2, 6) becomes line C, so every point on the line segment between (2, 6) and (4, 3)
becomes an optimal solution. For any value of PD > 750, the objective function line is even
steeper than line C, so (4, 3) becomes the only optimal solution.
Conclusion: The allowable range for PD is 0 ≤ PD ≤ 750, because (D, W) = (2, 6) remains
optimal over this range but not beyond. (When PD = 0 or PD = 750, there are multiple optimal
FIGURE 5.9
The two dashed lines
that pass through solid
constraint boundary lines
are the objective function
lines when PD (the unit
profit for doors) is at an
endpoint of its allowable
range, 0 ≤ PD ≤ 750,
since either line or any
objective function line in
between still yields (D, W)
= (2, 6) as an optimal
solution for the Wyndor
problem.
Production rate
for windows
W
8
(2, 6) is optimal for 0 ≤ PD ≤ 750
6
Line B
PD = 0 (Profit = 0D + 500W)
Line C
4
PD = 300 (Profit = 300D + 500W)
Feasible
region
2
PD = 750 (Profit = 750D + 500W)
Line A
0
2
4
6
Production rate for doors
D
162 Chapter Five What-If Analysis for Linear Programming
solutions, but (D, W) = (2, 6) still is one of them.) With the range this wide around the original
estimate of $300 (PD = 300) for the unit profit for doors, we can be quite confident of obtaining
the correct optimal solution for the true unit profit even though the discussion in Section 5.2
indicates that this estimate is fairly rough.
The sensitivity report also can be used to find the allowable range for the unit profit for
Wyndor’s other new product. In particular, let
​P​  W​​ = Unit profit for Wyndor’s new kind of window
​
​​ ​  ​
​​​
​ = Cell D4 in the spreadsheet
Referring to the Windows row of the sensitivity report (Figure 5.8), this row indicates that
the allowable decrease in PW is 300 (so PW ≥ 500 − 300 = 200) and the allowable increase
is 1E + 30. What is meant by 1E + 30? This is shorthand in Excel for 1030 (1 with 30 zeroes
after it). This tremendously huge number is used by Excel to represent infinity. Therefore, the
allowable range of PW is obtained from the sensitivity report as follows:
Current value of ​P​  W​​:
500
​
Allowable increase in ​P​  W​​: Unlimited
So ​P​  W​​ has no upper limit
​​
​
​
​
​
​  ​  ​
​ ​​​
Allowable decrease in P
​ ​  W​​: 300
So ​P​  W​​ ≥ 500 − 300 = 200
Allowable range:
​P​  W​​ ≥ 200 ​
A parameter is considered
sensitive if even a small
change in its value can
change the optimal solution.
Review
Questions
The allowable range is quite wide for both objective function coefficients. Thus, even
though PD = $300 and PW = $500 were only rough estimates of the true unit profit for the
doors and windows, respectively, we can still be confident that we have obtained the correct
optimal solution.
We are not always so lucky. For some linear programming problems, even a small change
in the value of certain coefficients in the objective function can change the optimal solution.
As first mentioned in Section 5.1, such coefficients are referred to as sensitive parameters.
The sensitivity report will immediately indicate which of the objective function coefficients
(if any) are sensitive parameters. These are parameters that have a small allowable increase
and/or a small allowable decrease. Hence, extra care should be taken to refine these estimates.
Once this has been done and the final version of the model has been solved, the allowable
ranges continue to serve an important purpose. As indicated in Section 5.1, the second benefit
of what-if analysis is that if conditions change after the study has been completed (a common
occurrence), what-if analysis leaves signposts that indicate (without solving the model again)
whether a resulting change in a parameter of the model changes the optimal solution. Thus, if
weeks, months, or even years later, the unit profit for one of Wyndor’s new products changes
substantially, its allowable range indicates immediately whether the old optimal product mix
still is the appropriate one to use. Being able to draw an affirmative conclusion without reconstructing and solving the revised model is extremely helpful for any linear programming problem, but especially so when the model is a large one.
1. What type of report can Analytic Solver generate to show how the optimal solution varies for
a defined range of values of a parameter cell?
2. What is meant by the allowable range for a coefficient in the objective function?
3. What is the significance if the true value for a coefficient in the objective function turns out to
be so different from its estimate that it lies outside its allowable range?
4. In Solver’s sensitivity report, what is the interpretation of the Objective Coefficient column?
The Allowable Increase column? The Allowable Decrease column?
5.4 THE EFFECT OF SIMULTANEOUS CHANGES IN OBJECTIVE
­FUNCTION COEFFICIENTS
The coefficients in the objective function typically represent quantities (e.g., unit profits) that
can only be estimated because of considerable uncertainty about what their true values will turn
out to be. The allowable ranges described in the preceding section deal with this uncertainty by
5.4 The Effect of Simultaneous Changes in Objective ­Function Coefficients 163
focusing on just one coefficient at a time. In effect, the allowable range for a particular coefficient assumes that the original estimates for all the other coefficients are completely accurate
so that this coefficient is the only one whose true value may differ from its original estimate.
In actuality, the estimates for all the coefficients (or at least more than one of them) may
be inaccurate simultaneously. The crucial question is whether this is likely to result in obtaining the wrong optimal solution. If so, greater care should be taken to refine these estimates
as much as possible, at least for the more crucial coefficients. On the other hand, if what-if
analysis reveals that the anticipated errors in estimating the coefficients are unlikely to affect
the optimal solution, then management can be reassured that the current linear programming
model and its results are providing appropriate guidance.
This section focuses on how to determine, without solving the problem again, whether the
optimal solution might change if certain changes occur simultaneously in the coefficients of
the objective function (due to their true values differing from their estimates). In the process,
we will address the second of Wyndor management’s what-if questions.
Question 2: What happens if the estimates of the unit profits of both of Wyndor’s new products
are inaccurate?
Using the Spreadsheet for This Analysis
Once again, a quick-and-easy way to address this kind of question is to simply try out different
estimates on the spreadsheet formulation of the model and see what happens each time after
running Solver again.
In this case, the optimal product mix indicated by the model is heavily weighted toward
producing the windows (6 per week) rather than the doors (only 2 per week). Since there is
equal enthusiasm for both new products, management is concerned about this imbalance.
Therefore, management has raised a what-if question. What would happen if the estimate of
the unit profit for the doors ($300) were too low and the corresponding estimate for the windows ($500) were too high? Management feels that the estimates could easily be off in these
directions. If this were the case, would this lead to a more balanced product mix being the
most profitable one?
This question can be answered in a matter of seconds simply by substituting new estimates of the unit profits in the original spreadsheet in Figure 5.1 and running Solver again.
­Figure 5.10 shows that new estimates of $450 for doors and $400 for windows causes no
change at all in the solution for the optimal product mix. (The total profit does change, but this
occurs only because of the changes in the unit profits.) Would even larger changes in the estimates of unit profits finally lead to a change in the optimal product mix? Figure 5.11 shows
that this does happen, yielding a relatively balanced product mix of (D, W) = (4, 3), when
estimates of $600 for doors and $300 for windows are used.
FIGURE 5.10
The revised Wyndor
problem where the estimates of the unit profits
for doors and windows
have been changed to
PD = $450 and PW =
$400, respectively, but
no change occurs in the
optimal solution.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$450
$400
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
≤
4
12
≤
12
18
≤
18
7
Plant 1
1
0
2
8
Plant 2
0
2
9
Plant 3
3
2
Doors
Windows
Total Profit
2
6
$3,300
10
11
12
Units Produced
164 Chapter Five What-If Analysis for Linear Programming
FIGURE 5.11
The revised Wyndor
problem where the estimates of the unit profits
for doors and windows
have been changed to
$600 and $300, respectively, which results in
a change in the optimal
solution.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$600
$300
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
4
≤
4
8
Plant 2
0
2
6
≤
12
9
Plant 3
3
2
18
≤
18
Doors
Windows
Total Profit
4
3
$3,300
10
11
12
Units Produced
Using a Two-Way Parameter Analysis Report Generated
by Analytic Solver for This Analysis
Using Analytic Solver, a two-way parameter analysis report provides a way of systematically investigating the effect if the estimates of both unit profits are inaccurate. This kind of
parameter analysis table shows the results in a single output cell for various trial values in
two parameter cells. Therefore, for example, it can be used to show how TotalProfit (G12) in
Figure 5.1 varies over a range of trial values in the two parameter cells, UnitProfitPerDoor
(C4) and UnitProfitPerWindow (D4). For each pair of trial values in these data cells, Solver is
called on to re-solve the problem.
To create such a two-way parameter analysis report for the Wyndor problem, both UnitProfitPerDoor (C4) and UnitProfitPerWindow (D4) need to be defined as parameter cells. In
turn, select cell C4 and D4, then choose Parameters > Optimization on the Analytic Solver ribbon, and then enter the range of trial values for each parameter cell (as was done in ­Figure 5.5
in the preceding section). For this example, UnitProfitPerDoor (C4) will be varied from $300
to $600 while UnitProfitPerWindow (D4) will be varied from $100 to $500.
Next, choose Reports > Optimization > Parameter Analysis on the Analytic Solver ribbon
to bring up the dialog box shown in Figure 5.12. For a two-way parameter analysis report, two
parameter cells are chosen, but only a single result can be shown. Under Parameters, clicking on
(>>) chooses both of the defined parameter cells, UnitProfitPerDoor and UnitProfitPer­Window.
Under Results, click on (<<) to clear out the list of cells on the right, click on the + next to
Objective to reveal the objective cell (TotalProfit), select TotalProfit, and then click on > to
move this cell to the right.
The next step is to choose Vary Two Selected Parameters Independently from the menu
toward the bottom of the window (as opposed to the other two options—Vary All Selected
Parameters Simultaneously or One At a Time). This will allow both parameter cells to be varied independently over their entire ranges. The number of different values of the first parameter cell and the second parameter cell to be shown in the parameter analysis report are entered
in Major Axis Points and Minor Axis Points, respectively. These values will be spread evenly
over the range of values specified in the parameter dialog box for each parameter cell. Therefore, choosing 4 and 5 for the respective number of values, as shown in Figure 5.12, will vary
UnitProfitPerDoor over the four values of $300, $400, $500, and $600 while simultaneously
varying UnitProfitPerWindow over the five values of $100, $200, $300, $400, and $500.
Clicking on the OK button generates the parameter analysis report shown in Figure 5.13.
The trial values for the respective parameter cells are listed in the first column and first row
of the table. For each combination of a trial value from the first column and from the first
row, Solver has solved for the value of the output cell of interest (the objective cell for this
example) and entered it into the corresponding column and row of the table.
5.4 The Effect of Simultaneous Changes in Objective ­Function Coefficients 165
FIGURE 5.12
The dialog box for the
parameter analysis report
specifies here that the
UnitProfitPerDoor and
UnitProfitPerWindow
parameter cells will be
varied and results from
the objective cell (TotalProfit) will be shown for
the Wyndor problem.
FIGURE 5.13
The parameter analysis
report that shows how
the optimal total profit
changes when systematically varying the estimate
of both the unit profit for
doors and the unit profit
for windows for the
Wyndor problem.
What-if analysis shows that
there is no need to refine
Wyndor’s estimates of the
unit profits for doors and
windows.
1
2
3
4
5
6
A
TotalProfit
UnitProfitPerDoor
$300
$400
$500
$600
B
UnitProfitPerWindow
$100
$1,500
$1,900
$2,300
$2,700
C
D
E
$200
$1,800
$2,200
$2,600
$3,000
$300
$2,400
$2,600
$2,900
$3,300
$400
$3,000
$3,200
$3,400
$3,600
F
$500
$3,600
$3,800
$4,000
$4,200
It also is possible to choose either DoorsProduced or WindowsProduced instead of TotalProfit as the result to show in the dialog box in Figure 5.12. A similar parameter analysis
report then could have been generated to show either the optimal number of doors to produce
or the optimal number of windows to produce for each combination of values for the unit
profits. These two parameter analysis reports are shown in Figure 5.14. The upper right-hand
corner (cell F3) of both reports, taken together, gives the optimal solution of (D, W) = (2, 6)
when using the original unit-profit estimates of $300 for doors and $500 for windows. ­Moving
down from this cell corresponds to increasing this profit estimate for doors, while moving to
the left amounts to decreasing the profit estimate for windows. Moving to the lower-left corner
of both tables reveals that (D, W) = (4, 3) becomes the optimal solution for these values of the
profit estimates. However, (D, W) = (2, 6) continues to be the optimal solution for all the cells
near F3. This indicates that the original estimates of unit profit would need to be very inaccurate indeed before the optimal product mix would change. Although the estimates are fairly
rough, management is confident that they are not that inaccurate. Therefore, there is no need to
expend the considerable effort that would be needed to refine the estimates.
166 Chapter Five What-If Analysis for Linear Programming
FIGURE 5.14
The pair of parameter
analysis reports that show
how the optimal number
of doors to produce (top
report) and the optimal
number of windows to
produce (bottom report)
change when systematically varying the estimate
of both the unit profit for
doors and the unit profit
for windows for the Wyndor problem.
1
2
3
4
5
6
A
DoorsProduced
UnitProfitPerDoor
$300
$400
$500
$600
B
UnitProfitPerWindow
$100
4
4
4
4
1
2
3
4
5
6
A
WindowsProduced
UnitProfitPerDoor
$300
$400
$500
$600
B
UnitProfitPerWindow
$100
3
3
3
3
C
D
E
$200
2
4
4
4
$300
2
2
4
4
$400
2
2
2
2
C
D
E
F
$200
6
3
3
3
$300
6
6
3
3
$400
6
6
6
6
$500
6
6
6
6
F
$500
2
2
2
2
At this point, it continues to appear that (D, W) = (2, 6) is the best product mix for initiating the production of the two new products (although additional what-if questions remain to
be addressed in subsequent sections). However, we also now know from Figure 5.14 that as
conditions change in the future, if the unit profits for both products change enough, it may
be advisable to change the product mix later. We still need to leave clear signposts behind to
signal when a future change in the product mix should be considered, as described next.
Gleaning Additional Information from the Sensitivity Report
The preceding section described how the data in the sensitivity report enable finding the
allowable range for an individual coefficient in the objective function when that coefficient is
the only one that changes from its original value. These same data (the allowable increase and
allowable decrease in each coefficient) also can be used to analyze the effect of simultaneous
changes in these coefficients. Here is how.
A sum ≤ 100 percent
guarantees that the original
optimal solution is still
optimal.
The 100 Percent Rule for Simultaneous Changes in Objective Function Coefficients: If simultaneous
changes are made in the coefficients of the objective function, calculate for each change the percentage of the allowable change (increase or decrease) for that coefficient to remain within its allowable
range. If the sum of the percentage changes does not exceed 100 percent, the original optimal solution definitely will still be optimal. (If the sum does exceed 100 percent, then we cannot be sure.)
This rule does not spell out what happens if the sum of the percentage changes does exceed
100 percent. The consequence depends on the directions of the changes in the coefficients.
Exceeding 100 percent may or may not change the optimal solution, but so long as 100 percent
is not exceeded, the original optimal solution definitely will still be optimal.
Keep in mind that we can safely use the entire allowable increase or decrease in a single
objective function coefficient only if none of the other coefficients have changed at all. With
simultaneous changes in the coefficients, we focus on the percentage of the allowable increase
or decrease that is being used for each coefficient.
To illustrate, consider the Wyndor problem again, along with the information provided by
the sensitivity report in Figure 5.8. Suppose conditions have changed after the initial study, and
the unit profit for doors (PD) has increased from $300 to $450 while the unit profit for windows
(PW) has decreased from $500 to $400. The calculations for the 100 percent rule then are
​P​  D​​: $300 → $450
​ ​
​ ​
450
−
300
​
Percentage of allowable increase = 100​​ ____________
​​​
​​ ​​​% = 33​1⁄3​%
(
450 )
​P
​
​
​​ W​  ​​:​  ​$500 → $400 ​  ​
​  ​​  ​  ​  ​  ​​ ​
​  ​  ​​ ​  ​  ​  ​​​
​​​​
500 − 400
1
____________
​
Percentage of allowable decrease = 100​​ ​​​
​​ ​​​% = 33​⁄3​%
(
300 )
¯
​
​
​
Sum = 66​
​  2⁄3​%​
5.4 The Effect of Simultaneous Changes in Objective ­Function Coefficients 167
Since the sum of the percentages does not exceed 100 percent, the original optimal solution
(D, W) = (2, 6) definitely is still optimal, just as we found earlier in Figure 5.10.
Now suppose conditions have changed even further, so PD has increased from $300 to $600
while PW has decreased from $500 to $300. The calculations for the 100 percent rule now are
​P​  D​​: $300 → $600
​ ​
​ ​
600
−
300
2
​
Percentage of allowable increase = 100​​ ____________
(​​​  450 ​​)​​​% = 66​⁄3​%
​P
​
​
​​ W​  ​​:​  ​$500 → $300 ​  ​
​  ​ ​​  ​  ​  ​​ ​
​  ​  ​​ ​  ​  ​  ​​​
​​ ​​
500
−
300
​
Percentage of allowable decrease = 100​​ ​​​ ____________
​​ ​​​% = 66​2⁄3​%
(
300 )
¯
​
​
​
Sum = 133​
​  1⁄3​%​
Since the sum of the percentages now exceeds 100 percent, the 100 percent rule says that
we can no longer guarantee that (D, W) = (2, 6) is still optimal. In fact, we found earlier in
both Figures 5.11 and 5.14 that the optimal solution has changed to (D, W) = (4, 3).
These results suggest how to find just where the optimal solution changes while PD is being
increased and PW is being decreased in this way. Since 100 percent is midway between 66 ​​2⁄3​​
percent and 133​​1⁄3​​percent, the sum of the percentage changes will equal 100 percent when
the values of PD and PW are midway between their values in the above cases. In particular,
PD = $525 is midway between $450 and $600 and PW = $350 is midway between $400 and
$300. The corresponding calculations for the 100 percent rule are
​P​  D​​: $300 → $525
​ ​
​ ​
525 − 300
Percentage of allowable increase = 100​​ ____________
​​​
​​ %
​​​ = 50%
(
450 )
​P
​
​​ ​  W​​:​  ​$500 → $300​  ​
​  ​  ​ ​ ​  ​​
​  ​  ​​  ​  ​ ​ ​​
500
−
350
​
Percentage of allowable decrease = 100​​ ____________
​​​
​​ ​​​% = 50%
(
300 )
​
​
​
Sum = ¯
​
100%​
​
Although the sum of the percentages equals 100 percent, the fact that it does not exceed
100 percent guarantees that (D, W) = (2, 6) is still optimal. Figure 5.15 shows graphically that
both (2, 6) and (4, 3) are now optimal, as well as all the points on the line segment connecting
these two points. However, if PD and PW were to be changed any further from their original
values (so that the sum of the percentages exceeds 100 percent), the objective function line
would be rotated so far toward the vertical that (D, W) = (4, 3) would become the only optimal
solution.
At the same time, keep in mind that having the sum of the percentages of allowable changes
exceed 100 percent does not automatically mean that the optimal solution will change. For
example, suppose that the estimates of both unit profits are halved. The resulting calculations
for the 100 percent rule are
​P​  D​​: $300 → $150
​ ​
​ ​
300 − 150
Percentage of allowable decrease = 100​​ ____________
​​​
​​ %
​​​ = 50%
(
300 )
​P
​  W​​:​  ​$500 → $250​  ​
​
​​
​  ​ ​ ​  ​​
​  ​​  ​  ​ ​ ​​
500
−
250
​
Percentage of allowable decrease = 100​​ ____________
​​​
​​ ​​​% = 83%
(
300 )
​
​
​
Sum = ¯
133%​
​
​
Here is an example where
the original optimal solution is still optimal even
though the sum exceeds
100 percent.
Even though this sum exceeds 100 percent, Figure 5.16 shows that the original optimal solution is still optimal. In fact, the objective function line has the same slope as the original objective function line (the solid line in Figure 5.9). This happens whenever proportional changes
are made to all the unit profits, which will automatically lead to the same optimal solution.
168 Chapter Five What-If Analysis for Linear Programming
FIGURE 5.15
When the estimates of
the unit profits for doors
and windows change to
PD = $525 and PW =
$350, which lies at the
edge of what is allowed by
the 100 percent rule, the
graphical method shows
that (D, W) = (2, 6) still
is an optimal solution, but
now every other point on
the line segment between
this solution and (4, 3)
also is optimal.
Production rate
for windows
W
10
8
Objective function line now is
Profit = $3,150 = 525D + 350W
since PD = $525 and PW = $350
(2, 6)
6
Entire line segment is optimal
4
Feasible
region
(4, 3)
2
0
FIGURE 5.16
When the estimates of
the unit profits for doors
and windows change to
PD = $150 and PW = $250
(half their original
values), the graphical
method shows that the
optimal solution still is
(D, W) = (2, 6), even
though the 100 percent
rule says that the optimal
solution might change.
2
4
Production rate for doors
Production rate
for windows
Profit = $1,800 = 150D + 250W
6
8
D
W
8
Optimal solution
6
(2, 6)
4
Feasible
region
2
0
2
4
Production rate for doors
6
8
D
Comparisons
You now have seen three approaches to investigating what happens if simultaneous changes
occur in the coefficients of the objective function: (1) try out changes directly on a spreadsheet, (2) use a two-way parameter analysis report and (3) apply the 100 percent rule.
5.5 The Effect of Single Changes in a Constraint 169
The spreadsheet approach is a good place to start, especially for less experienced modelers, because it is simple and quick. If you are only interested in checking one specific set of
changes in the coefficients, you can immediately see what happens after making the changes
in the spreadsheet.
More often, there will be numerous possibilities for what the true values of the coefficients
will turn out to be, because of uncertainty in the original estimates of these coefficients. The
parameter analysis report is useful for systematically checking a variety of possible changes
in one or two objective function coefficients. Trying out representative possibilities on the
spreadsheet may provide all the insight that is needed. Perhaps the optimal solution for the
original model will remain optimal over nearly all these possibilities, so this solution can be
confidently used. Or perhaps it will become clear that the original estimates need to be refined
before selecting a solution.
When the spreadsheet approach and/or parameter analysis report does not provide a clear
conclusion, the 100 percent rule can usefully complement this approach in the following ways:
• The 100 percent rule can be used to determine just how large the changes in the objective
function coefficients need to be before the original optimal solution may no longer be
optimal.
• When the model has a large number of decision variables (as is common for real problems), it may become impractical to use the spreadsheet approach to systematically try out a
variety of simultaneous changes in many or all of the coefficients in the objective function
because of the huge number of representative possibilities. The parameter analysis report
can only be used to systematically check possible changes in—at most—two coefficients
at a time. However, by dividing each coefficient’s allowable increase or allowable decrease
by the number of decision variables, the 100 percent rule immediately indicates how much
each coefficient can be safely changed without invalidating the current optimal solution.
• After completing the study, if conditions change in the future that cause some or all of
the coefficients in the objective function to change, the 100 percent rule quickly indicates whether the original optimal solution must remain optimal. If the answer is affirmative, there is no need to take all the time that may be required to reconstruct the (revised)
spreadsheet model. The time saved can be very substantial for large models.
Review
Questions
5.5
1. How many result cells can be selected for display in Analytic Solver’s two-way parameter
analysis report?
2. In the 100 percent rule for simultaneous changes in objective function coefficients, what are
the percentage changes that are being considered?
3. In this 100 percent rule, if the sum of the percentage changes does not exceed 100 percent,
what does this say about the original optimal solution?
4. In this 100 percent rule, if the sum of the percentage changes exceeds 100 percent, does this
mean that the original optimal solution is no longer optimal?
THE EFFECT OF SINGLE CHANGES IN A CONSTRAINT
When the right-hand sides
represent managerial policy
decisions, what-if analysis
provides guidance regarding the effect of altering
these decisions.
We now turn our focus from the coefficients in the objective function to the effect of changing
the functional constraints. The changes might occur either in the coefficients on the left-hand
sides of the constraints or in the values of the right-hand sides.
We might be interested in the effect of such changes for the same reason we are interested
in this effect for objective function coefficients, namely, that these parameters of the model
are only estimates of quantities that cannot be determined precisely at this time so we want to
determine the effect if these estimates are inaccurate.
However, a more common reason for this interest is the one discussed at the end of
­Section 5.1, namely, that the right-hand sides of the functional constraints may well represent managerial policy decisions rather than quantities that are largely outside the control of
management. Therefore, after the model has been solved, management will want to analyze
the effect of altering these policy decisions in a variety of ways to see if these decisions can
170 Chapter Five What-If Analysis for Linear Programming
be improved. What-if analysis provides valuable guidance to management in determining the
effect of altering these policy decisions. (Recall that this was cited as the third benefit of
what-if analysis in Section 5.1.)
This section describes how to perform what-if analysis when making changes in just one
spot (a coefficient or a right-hand side) of a single constraint. The next section then will deal
with simultaneous changes in the constraints.
The procedure for determining the effect if a single change is made in a constraint is the
same regardless of whether the change is in a coefficient on the left-hand side or in the value
on the right-hand side. (The one exception is that the Solver sensitivity report provides information about changes in the right-hand side but does not do so for the left-hand side.) Therefore, we will illustrate the procedure by making changes in a right-hand side.
In particular, we return to the Wyndor case study to address the third what-if question
posed by Wyndor management in Section 5.2.
Question 3: What happens if a change is made in the number of hours of production time per
week being made available to Wyndor’s new products in one of the plants?
The number of hours available in each plant is the value of the right-hand side for the corresponding constraint, so we want to investigate the effect of changing this right-hand side for
one of the plants. With the original optimal solution, (D, W) = (2, 6), only 2 of the 4 available
hours in Plant 1 are used, so changing this number of available hours (barring a large decrease)
would have no effect on either the optimal solution or the resulting total profit from the two
new products. However, it is unclear what would happen if the number of available hours in
either Plant 2 or Plant 3 were to be changed. Let’s start with Plant 2.
Using the Spreadsheet for This Analysis
Referring back to Section 5.2, Figure 5.1 shows the spreadsheet model for the original Wyndor
problem before beginning what-if analysis. The optimal solution is (D, W) = (2, 6) with a total
profit of $3,600 per week from the two new products. Cell G8 shows that 12 hours of production time per week are being made available for the new products in Plant 2.
To see what happens if a specific change is made in this number of hours, all you need to do
is substitute the new number in cell G8 and run Solver again. For example, Figure 5.17 shows
the result if the number of hours is increased from 12 to 13. The corresponding optimal solution in C12:D12 gives a total profit of $3,750. Thus, the resulting change in profit would be
Incremental profit = $3,750 − $3,600
​ ​​
​  ​
​​​
​
= $150
Since this increase in profit is obtained by adding just one more hour in Plant 2, it would
be interesting to see the effect of adding several more hours. Figure 5.18 shows the effect of
FIGURE 5.17
The revised Wyndor
problem where the hours
available in Plant 2 per
week have been increased
from 12 (as in Figure 5.1)
to 13, which results in an
increase of $150 in the
total profit per week from
the two new products.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
1.66667
≤
4
8
Plant 2
0
2
13
≤
13
9
Plant 3
3
2
18
≤
18
Doors
Windows
Total Profit
1.667
6.5
$3,750
10
11
12
Units Produced
5.5 The Effect of Single Changes in a Constraint 171
FIGURE 5.18
A further revision of
the Wyndor problem in
Figure 5.17 to further
increase the hours available in Plant 2 from 13
to 18, which results in a
further increase in total
profit of $750 (which is
the $150 per hour added
in Plant 2).
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
Unit Profit
4
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
0
≤
4
8
Plant 2
0
2
18
≤
18
9
Plant 3
3
2
18
≤
18
Doors
Windows
Total Profit
0
9
$4,500
10
11
12
Units Produced
adding five more hours. Comparing Figure 5.18 to Figure 5.17, the additional profit from
providing five more hours would be
So far, each additional hour
provided in Plant 2 adds
$150 to profit.
Now management needs
to consider the trade-off
between adding production
time for the new products
and decreasing it for other
products.
Incremental profit = $4,500 − $3,750
​ ​​​
​  =​ ​  $750 from adding​ 5 hours​
​​
​
= $150 per hour added
Would adding even more hours increase profit even further? Figure 5.19 shows what would
happen if a total of 20 hours per week were made available to the new products in Plant 2. Both
the optimal solution and the total profit are the same as in Figure 5.18, so increasing from 18 to
20 hours would not help. (The reason is that the 18 hours available in Plant 3 prevent producing
more than 9 windows per week, so only 18 hours can be used in Plant 2.) Thus, it appears that
18 hours is the maximum that should be considered for Plant 2.
However, the fact that the total profit from the two new products can be increased substantially by increasing the number of hours per week made available to the new products
from 12 to 18 does not mean that these additional hours should be provided automatically.
The production time made available for these two new products can be increased only if it is
decreased for other products. Therefore, management will need to assess the disadvantages
of decreasing the production time for any other products (including both lost profit and less
tangible disadvantages) before deciding whether to increase the production time for the new
products. This analysis also might lead to decreasing the production time made available to
the two new products in one or more of the plants.
FIGURE 5.19
A further revision of
the Wyndor problem in
Figure 5.18 to further
increase the hours
available in Plant 2 from
18 to 20, which results in
no change in total profit
because the optimal solution cannot make use of
these additional hours.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
0
≤
4
8
Plant 2
0
2
18
≤
20
9
Plant 3
3
2
18
≤
18
Doors
Windows
Total Profit
0
9
$4,500
10
11
12
Units Produced
172 Chapter Five What-If Analysis for Linear Programming
Using a Parameter Analysis Report Generated by Analytic
Solver for This Analysis
In general, the shadow price
for a constraint reveals the
rate at which the objective
cell can be increased by
increasing the right-hand
side of that constraint. This
remains valid as long as the
right-hand side is within its
allowable range.
This allowable range
focuses on a right-hand
side and the corresponding
shadow price. In contrast
to the allowable range
for objective function
coefficients described in
Section 5.3, it does not
indicate whether the original solution is still optimal,
just whether the shadow
price remains valid.
Using Analytic Solver, a parameter analysis report can be used to systematically determine
the effect of making various changes in one of the parameters in a constraint. For example,
Figure 5.20 displays a parameter analysis report (obtained by following the same procedure
used to generate a parameter analysis report in Section 5.3) to show how the changing cells
and total profit change as the number of available hours in Plant 2 range between 4 and 20.
An interesting pattern is apparent in the incremental profit column. Starting at 12 hours
available at Plant 2 (the current allotment), each additional hour allocated yields an additional
$150 in profit (up to making 18 hours available). Similarly, if hours are taken away from
Plant 2, each hour lost causes a loss of $150 profit (down to making six hours available). This
rate of change in the profit for increases or decreases in the right-hand side of a constraint is
known as the shadow price.
Given an optimal solution and the corresponding value of the objective function for a linear
programming model, the shadow price for a functional constraint is the rate at which the value
of the objective function can be increased by increasing the right-hand side of the constraint by
a small amount.
However, the shadow price of $150 for the Plant 2 constraint is valid only within a range of
values near 12 (in particular, between 6 hours and 18 hours). If the number of available hours
is increased beyond 18 hours, then the incremental profit drops to zero. If the available hours
are reduced below six hours, then profit drops at a faster rate of $250 per hour. Therefore, letting RHS denote the value of the right-hand side, the shadow price of $150 is valid for
​6 ≤ RHS ≤ 18​
This range is known as the allowable range for the right-hand side (or just allowable
range for short).
The allowable range for the right-hand side of a functional constraint is the range of values for
this right-hand side over which this constraint’s shadow price remains valid.
Unlike the allowable range for objective function coefficients described in Section 5.3, a
change that is within the allowable range for the right-hand side does not mean the original
FIGURE 5.20
The parameter analysis report that shows the effect of varying the number of hours of production time being made available per week
in Plant 2 for Wyndor’s new products.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
A
HoursAvailableInPlant2
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
B
DoorsProduced
4
4
4
3.666666667
3.333333333
3
2.666666667
2.333333333
2
1.666666667
1.333333333
1
0.666666667
0.333333333
0
0
0
C
WindowsProduced
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9
9
D
TotalProfit
$2,200
$2,450
$2,700
$2,850
$3,000
$3,150
$3,300
$3,450
$3,600
$3,750
$3,900
$4,050
$4,200
$4,350
$4,500
$4,500
$4,500
E
Incremental Profit
$250
$250
$150
$150
$150
$150
$150
$150
$150
$150
$150
$150
$150
$150
$0
$0
5.5 The Effect of Single Changes in a Constraint 173
solution is still optimal. In fact, any time the shadow price is not zero, a change to a right-hand
side leads to a change in the optimal solution. The shadow price indicates how much the value
of the objective function will change as the optimal solution changes.
Using the Sensitivity Report to Obtain the Key Information
The shadow prices reveal
the relationship between
profit and the amount of
production time made
available in the plants.
FIGURE 5.21
The complete sensitivity
report generated by Solver
for the original Wyndor
problem as formulated in
Figure 5.1.
As illustrated earlier, it is straightforward to use a parameter analysis report to calculate the
shadow price for a functional constraint, as well as to find (or at least closely approximate)
the allowable range for the right-hand side of this constraint over which the shadow price
remains valid. However, this same information also can be obtained immediately from Solver’s sensitivity report for all the functional constraints. Figure 5.21 shows the full sensitivity
report provided by Solver for the original Wyndor problem after obtaining the optimal solution given in Figure 5.1. The top half is the part already shown in Figure 5.8 for finding allowable ranges for the objective function coefficients. The bottom half focuses on the functional
constraints, including providing the shadow prices for these constraints in the fourth column.
The first three columns remind us that (1) the output cells for these constraints in Figure 5.1
are cells E7 to E9, (2) these cells give the number of production hours used per week in the
three plants, and (3) the final values in these cells are 2, 12, and 18 (as shown in column E of
Figure 5.1). (We will discuss the last three columns a little later.)
The shadow price given in the fourth column for each constraint tells us how much the
value of the objective function [objective cell (G12) in Figure 5.1] would increase if the righthand side of that constraint (cell G7, G8, or G9) were to be increased by 1. Conversely, it also
tells us how much the value of the objective function would decrease if the right-hand side
were to be decreased by 1. The shadow price for the Plant 1 constraint is 0, because this plant
already is using less hours (2) than are available (4) so there would be no benefit to making
an additional hour available. However, Plants 2 and 3 are using all the hours available to them
for the two new products (with the product mix given by the changing cells). Thus, it is not
surprising that the shadow prices indicate that the objective cell would increase if the hours
available in either Plant 2 or Plant 3 were to be increased.
To express this information in the language of management, the value of the objective
function for this problem [objective cell (G12) in Figure 5.1] represents the total profit in
dollars per week from the two new products under consideration. The right-hand side of each
functional constraint represents the number of hours of production time being made available per week for these products in the plant that corresponds to this constraint. Therefore,
the shadow price for a functional constraint informs management as to how much the total
profit from the two new products could be increased for each additional hour of production
time made available to these products per week in the corresponding plant. Conversely, the
shadow price indicates how much this profit would decrease for each reduction of an hour of
production time in that plant. This interpretation of the shadow price remains valid as long as
the change in the number of hours of production time is not very large.
Specifically, this interpretation of the shadow price remains valid as long as the number
of hours of production time remains within its allowable range. Solver’s sensitivity report
Variable Cells
Name
Final
Value
DoorsProduced
WindowsProduced
2
6
Cell
$C$12
$D$12
Reduced
Cost
0
0
Objective
Coefficient
300
500
Allowable
Increase
450
IE + 30
Allowable
Decrease
300
300
Constraints
Cell
Name
Final
Value
$E$7
$E$8
$E$9
Plant 1 Used
Plant 2 Used
Plant 3 Used
2
12
18
Shadow
Price
0
150
100
Constraint
R. H. Side
4
12
18
Allowable
Increase
Allowable
Decrease
IE + 30
6
6
2
6
6
174 Chapter Five What-If Analysis for Linear Programming
Here is how to find the
allowable ranges for the
right-hand sides from the
sensitivity report.
provides all the data needed to identify the allowable range of each functional constraint.
Refer back to the bottom of this report given in Figure 5.21. The final three columns enable
calculation of this range. The “Constraint R. H. Side” column indicates the original value of
the right-hand side before any change is made. Adding the number in the “Allowable Increase”
column to this original value then gives the upper endpoint of the allowable range. Similarly,
subtracting the number in the “Allowable Decrease” column from this original value gives the
lower endpoint. Using the fact that 1E + 30 represents infinity (∞), these calculations of the
allowable ranges are shown below, where a subscript has been added to each RHS to identify
the constraint involved.
​​​Plant 1 constraint:​  4 − 2 ≤ RHS1 ≤ 4 + ∞,​  so 2 ≤ RHS​1​​ (no upper limit)​
Plant 2 constraint:​  12 − 6 ≤ RHS2 ≤ 12 + 6,​  so 6 ≤ RHS2​​ ≤ 18​
Plant 3 constraint:​  18 − 6 ≤ RHS3 ≤ 18 + 6,​  so 12 ≤ RHS​3​​ ≤ 24 ​​​
In the case of the Plant 2 constraint, Figure 5.22 provides graphical insight into why 6 ≤ RHS
≤ 18 is the range of validity for the shadow price. The optimal solution for the original problem,
(D, W) = (2, 6), lies at the intersection of line B and line C. The equation for line B is 2W = 12
because this is the constraint boundary line for the Plant 2 constraint (2W ≤ 12). However, if the
value of this right-hand side (RHS2 = 12) is changed, line B will either shift upward (for a larger
value of RHS2) or downward (for a smaller value of RHS2). As line B shifts, the boundary of the
feasible region shifts accordingly and the optimal solution continues to lie at the intersection of
the shifted line B and line C—provided the shift in line B is not so large that this intersection is no
longer feasible. Each time RHS2 is increased (or decreased) by 1, this intersection shifts enough
to increase (or decrease) Profit by the amount of the shadow price ($150). Figure 5.22 indicates
that this intersection remains feasible (and so optimal) as RHS2 increases from 12 to 18, because
the feasible region expands upward as line B shifts upward. However, for values of RHS2 larger
than 18, this intersection is no longer feasible because it gives a negative value of D (the production rate for doors). Thus, each increase of 1 above 18 no longer increases Profit by the amount
FIGURE 5.22
A graphical interpretation
of the allowable range,
6 ≤ RHS2 ≤ 18, for
the right-hand side of
Wyndor’s Plant 2
constraint.
W
Production rate
for windows
10
(0, 9)
2W = 18
Profit = 300(0) + 500(9) = $4,500
2W = 12
Profit = 300(2) + 500(6) = $3,600
8
6
4
2
Line B
(2, 6)
Feasible
region for
original
problem
(4, 3)
2W = 6
Profit = 300(4) + 500(3) = $2,700
Line C (3D + 2W = 18)
Line A (D = 4)
0
2
4
6
Production rate for doors
8
D
5.6 The Effect of Simultaneous Changes in the Constraints 175
of the shadow price. Similarly, as RHS2 decreases from 12 to 6, this intersection remains feasible
(and so optimal) as line B shifts down accordingly. However, for values of RHS2 less than 6, this
intersection is no longer feasible because it violates the Plant 1 constraint (D ≤ 4) whose boundary line is line A. Hence, each decrease of 1 below 6 no longer decreases Profit by the amount
of the shadow price. Consequently, 6 ≤ RHS ≤ 18 is the allowable range over which the shadow
price is valid.
Summary
Recall again that the right-hand side of each functional constraint for the Wyndor problem
represents the number of hours of production time per week in the corresponding plant that is
being made available to the two new products. The shadow price for each constraint reveals
how much the total profit from these new products would increase for each additional hour
of production time made available in the corresponding plant for these products. This interpretation of the shadow price remains valid as long as the number of hours remains within
its allowable range. Therefore, each shadow price can be applied by management to evaluate
a change in its original decision regarding the number of hours as long as the new number
is within the corresponding allowable range. This evaluation also would need to take into
account how the change in the number of hours made available to the new products would
impact the production rates and profits for the company’s other products.
Review
Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
5.6
Why might it be of interest to investigate the effect of making changes in a functional constraint?
Why might it be possible to alter the right-hand side of a functional constraint?
What is meant by a shadow price?
How can a shadow price be found by using the spreadsheet? By using a parameter analysis
report? By using Solver’s sensitivity report?
Why are shadow prices of interest to managers?
Can shadow prices be used to determine the effect of decreasing rather than increasing the
right-hand side of a functional constraint?
What does a shadow price of 0 tell a manager?
Which columns of Solver’s sensitivity report are used to find the allowable range for the righthand side of a functional constraint?
Why are these allowable ranges of interest to managers?
THE EFFECT OF SIMULTANEOUS CHANGES IN THE CONSTRAINTS
Managerial policy decisions
involving right-hand sides
frequently are interrelated,
so changes in these decisions should be considered
simultaneously.
The preceding section described how to perform what-if analysis to investigate the effect of
changes in a single spot of a constraint. We now turn our consideration to the effect of simultaneous changes in the constraints.
The need to consider these simultaneous changes arises frequently. There may be considerable uncertainty about the estimates for a number of the parameters in the functional
constraints, so questions will arise as to the effect if the true values of the parameters simultaneously deviate significantly from the estimates. Since the right-hand sides of the constraints
often represent managerial policy decisions, questions will arise about what would happen if
some of these decisions were to be changed. These decisions frequently are interrelated and
so need to be considered simultaneously.
We outline next how the usual three methods for performing what-if analysis can be
applied to considering simultaneous changes in the constraints. The third one (using Solver’s
sensitivity report) is only helpful for changing right-hand sides. For the first two (using the
spreadsheet and using a parameter analysis report), the procedure is the same regardless of
whether the changes are in the coefficients on the left-hand sides or in the right-hand sides of
the constraints (or both). Since changing the right-hand sides is the more important case, we
will focus on this case.
In particular, we now will deal with the fourth of Wyndor management’s what-if questions.
Question 4: What happens if simultaneous changes are made in the number of hours of production
time per week being made available to Wyndor’s new products in all the plants?
176 Chapter Five What-If Analysis for Linear Programming
In particular, after seeing that the Plant 2 constraint has the largest shadow price (150),
versus a shadow price of 100 for the Plant 3 constraint, management now is interested in
exploring a specific type of simultaneous change in these production hours. By shifting the
production of one of the company’s current products from Plant 2 to Plant 3, it is possible to
increase the number of production hours available to the new products in Plant 2 by decreasing the number of production hours available in Plant 3 by the same amount. Management
wonders what would happen if these simultaneous changes in production hours were made.
Using the Spreadsheet for This Analysis
According to the shadow prices, the effect of shifting one hour of production time per week
from Plant 3 to Plant 2 would be as follows.
​RHS​  2​​: 12 → 13 Change in total profit = Shadow price = $150
100​​​​
​
​​​RHS​  3​​:​  ​ 18 → 17   ​
​  Change in total profit = − Shadow price = −​_​
​
​
Net increase in total profit = $50
We now are checking to see
whether the shadow prices
remain valid for evaluating
specific simultaneous
changes in the right-hand
sides.
However, we don’t know if these shadow prices remain valid if both right-hand sides are
changed by this amount.
A quick way to check this is to substitute the new right-hand sides into the original spreadsheet in Figure 5.1 and run Solver again. The resulting spreadsheet in Figure 5.23 shows that
the net increase in total profit (from $3,600 to $3,650) is indeed $50, so the shadow prices are
valid for these particular simultaneous changes in right-hand sides.
How long will these shadow prices remain valid if we continue shifting production hours
from Plant 3 to Plant 2? We could continue checking this by substituting other combinations
of right-hand sides into the spreadsheet and re-solving each time. However, a more systematic
way of doing this is to use a parameter analysis report, as described next.
Using a Parameter Analysis Report Generated by Analytic
Solver for This Analysis
Since it can become tedious, or even impractical, to use the spreadsheet to investigate a large
number of simultaneous changes in the right-hand sides, let us see how Analytic Solver’s
parameter analysis report can be used to do this analysis more systematically.
We could use a two-way parameter analysis report to investigate how the profit and optimal
production rates vary for different combinations of the number of hours available in Plant 2
and Plant 3. However, in this case we aren’t interested in all combinations of hours in the two
plants, but rather only those combinations that involve a simple shifting of available hours
from Plant 3 to Plant 2. For this analysis, we will see that a one-way parameter analysis table
is sufficient.
FIGURE 5.23
The revised Wyndor
problem where column
G in Figure 5.1 has been
changed by shifting one
of the hours available in
Plant 3 to Plant 2 and then
re-solving.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
1.33333
4
8
Plant 2
0
2
13
13
9
Plant 3
3
2
17
17
Doors
Windows
Total Profit
1.333
6.5
$3,650
10
11
12
Units Produced
5.6 The Effect of Simultaneous Changes in the Constraints 177
By entering a formula into
one data cell in terms of
another one, a one-way
parameter analysis report
is able to investigate interrelated trial values in both
data cells.
For each hour reduced in Plant 3, an additional hour is made available in Plant 2. Thus, the
number of available hours in Plant 2 is a function of the number of available hours in Plant 3.
In particular, since there are 30 total hours available at the two plants (RHS2 + RHS3 = 30),
the number of available hours in Plant 2 (RHS2) is
​​RHS​  2​​ = 30 − ​RHS3​​​​
Figure 5.24 shows the Wyndor Glass Co. spreadsheet with the data cell for the number of
available hours in Plant 2 replaced by the above formula. Because of this formula, whenever
the number of available hours in Plant 3 is reduced, the number of available hours in Plant 2
will automatically increase by the same amount. Now a one-way parameter analysis report can
be used to investigate various numbers of available hours in Plant 3 (with the corresponding
automatic adjustment made to the available hours at Plant 2). HoursAvailableInPlant3 (G9)
is specified as a parameter cell with a range of trial values from 18 down to 12. A parameter
analysis report is then generated in Figure 5.25 to show how DoorsProduced (C12), WindowsProduced (D12), and TotalProfit (G12) vary as HoursAvailableInPlant3 (G9) varies from 18
down to 12. A column was added to the left of the parameter analysis report to show how
FIGURE 5.24
By inserting a formula into cell G8 that keeps the total number of hours available in Plant 2 and Plant 3 equal to 30, it will be
possible to generate a one-way parameter analysis report (see Figure 5.25) that shows the effect of shifting more and more of the
hours available from Plant 3 to Plant 2.
A
1
B
C
D
E
F
G
H
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
2
4
Total (Plants 2 & 3)
8
Plant 2
0
2
12
12
30
9
Plant 3
3
2
18
18
Doors
Windows
Total Profit
2
6
$3,600
10
11
12
Units Produced
G
H
5
Hours
6
Available
7
4
Total (Plants 2 & 3)
8
=H8-G9
30
9
18
FIGURE 5.25
The parameter analysis report that shows the effect of shifting more and more of the hours available from Plant 3 to Plant 2 for the
Wyndor problem.
1
2
3
4
5
6
7
8
A
HoursAvailableInPlant2
12
13
14
15
16
17
18
B
HoursAvailableInPlant3
18
17
16
15
14
13
12
C
DoorsProduced
2
1.333
0.667
0
0
0
0
D
WindowsProduced
6
6.5
7
7.5
7
6.5
6
E
TotalProfit
$3,600
$3,650
$3,700
$3,750
$3,500
$3,250
$3,000
F
Incremental Profit
$50
$50
$50
-$250
-$250
-$250
178 Chapter Five What-If Analysis for Linear Programming
HoursAvailableInPlant2 varies correspondingly with the different values of HoursAvailableInPlant3. Also, we have calculated the incremental profit (in column F) for each hour shifted
from Plant 3 to Plant 2.
Again there is a pattern to the incremental profit. For each hour shifted from Plant 3 to
Plant 2 (up to 3 hours), an additional profit of $50 is achieved. However, if more than 3 hours
are shifted, the incremental profit per hour shifted becomes −$250. Thus, it appears worthwhile to shift up to 3 available hours from Plant 3 to Plant 2, but no more.
Although a one-way parameter analysis report is limited to enumerating trial values for
only one data cell, you have just seen how such a parameter analysis report still can systematically investigate a large number of simultaneous changes in two data cells by entering a
formula for the second data cell in terms of the first one. The two data cells considered above
happened to be right-hand sides of constraints, but either or both could have been coefficients
on the left-hand side instead. It is even possible to enter formulas for multiple data cells in
terms of the one whose trial values are being enumerated. Furthermore, by using a two-way
parameter analysis report, trial values can be enumerated simultaneously for two data cells,
with the possibility of entering formulas for additional data cells in terms of these two.
Gleaning Additional Information from the Sensitivity Report
This 100 percent rule
reveals whether the simultaneous changes in the
right-hand sides are small
enough to guarantee that
the shadow prices are still
valid.
Despite the versatility of a parameter analysis report, it cannot handle a number of important
cases. The most important one is where management wants to explore various possibilities for
changing its policy decisions that correspond to changing several right-hand sides simultaneously in a variety of ways. Although the spreadsheet can be used to see the effect of any combination of simultaneous changes, it can take an exorbitant amount of time to systematically
investigate a large number of simultaneous changes in right-hand sides in this way. Fortunately,
Solver’s sensitivity report provides valuable information for guiding such an investigation. In
particular, there is a 100 percent rule (analogous to the one presented in Section 5.4) that uses
this information to conduct this kind of investigation.
Recall that the 100 percent rule described in Section 5.4 is used to investigate simultaneous
changes in objective function coefficients. The new 100 percent rule presented next investigates simultaneous changes in right-hand sides in a similar way.
The data needed to apply the new 100 percent rule for the Wyndor problem are given by the
last three columns in the bottom part of the sensitivity report in Figure 5.21. Keep in mind that
we can safely use the entire allowable decrease or increase from the current value of a right-hand
side only if none of the other right-hand sides are changed at all. With simultaneous changes in
the right-hand sides, we focus for each change on the percentage of the allowable decrease or
increase that is being used. As detailed next, the 100 percent rule basically says that we can safely
make the simultaneous changes only if the sum of these percentages does not exceed 100 percent.
The 100 Percent Rule for Simultaneous Changes in Right-Hand Sides: The shadow prices
remain valid for predicting the effect of simultaneously changing the right-hand sides of some of
the functional constraints as long as the changes are not too large. To check whether the changes
are small enough, calculate for each change the percentage of the allowable change (decrease or
increase) for that right-hand side to remain within its allowable range. If the sum of the percentage changes does not exceed 100 percent, the shadow prices definitely will still be valid. (If the
sum does exceed 100 percent, then we cannot be sure.)
To illustrate this rule, consider again the simultaneous changes (shifting one hour of production time per week from Plant 3 to Plant 2) that led to Figure 5.23. The calculations for the
100 percent rule in this case are
​RHS​  2​​: 12 → 13
​ ​
​ ​
13 − 12
2
Percentage of allowable increase = 100​​ _
(​​​  6 ​​)​​​ = 16​⁄3​%
​R
​
​​ HS​  3​​:​  ​18 → 17​  ​
​  ​  ​ ​ ​  ​ ​
​  ​  ​​  ​  ​​
​​​
18
−
17
2
​
Percentage of allowable decrease = 100​​ _
(​​​  6 ​​)​​​ = 16​⁄3​%
¯
​
​
​
Sum = 33​
​  1⁄3​%​
​
5.7 Robust Optimization 179
Since the sum of 33​​1⁄3​​percent is less than 100 percent, the shadow prices definitely are valid
for predicting the effect of these changes, as was illustrated with Figure 5.23.
The fact that 33​​1⁄3​​percent is one-third of 100 percent suggests that the changes can be three
times as large as above without invalidating the shadow prices. To check this, let us apply the
100 percent rule with these larger changes.
​RHS​  2​​: 12 → 15
​ ​
​ ​
15 − 12
Percentage of allowable increase = ​​ ​​ ​ _
​​ %
​​​ = 50%
(
6 )
HS​  3​​:​  ​18 → 15​  ​
​​
​ ​R
​  ​ ​ ​  ​​
​  ​​  ​  ​ ​ ​​
18
−
15
​
Percentage of allowable decrease = ​​ ​​ ​ _
​​ ​​​% = 50%
(
6 )
​
​
​
Sum = ¯
​
100%​
​
Because the sum does not exceed 100 percent, the shadow prices are still valid, but these are
the largest changes in the right-hand sides that can provide this guarantee. In fact, Figure 5.25
demonstrated that the shadow prices become invalid for larger changes.
Review
Questions
5.7
1. Why might it be of interest to investigate the effect of making simultaneous changes in the
functional constraints?
2. How can the spreadsheet be used to investigate simultaneous changes in the functional
constraints?
3. What are the capabilities of a parameter analysis report for investigating simultaneous changes
in the functional constraints?
4. Why might a manager be interested in considering simultaneous changes in right-hand sides?
5. What is the 100 percent rule for simultaneous changes in right-hand sides?
6. What are the data needed to apply the 100 percent rule for simultaneous changes in righthand sides?
7. What is guaranteed if the sum of the percentages of allowable changes in the right-hand sides
does not exceed 100 percent?
8. What is the conclusion if the sum of the percentages of allowable changes in the right-hand
sides does exceed 100 percent?
ROBUST OPTIMIZATION
As described in the preceding sections, what-if analysis provides an important way of dealing
with uncertainty about the true values of the parameters in a linear programming model. One
main purpose of what-if analysis is to identify each of the sensitive parameters, namely, each
of the parameters where even a small change in its value can change the optimal solution. This
is valuable information since these are the parameters that need to be estimated with special
care to minimize the risk of obtaining an erroneous optimal solution.
Unfortunately, it often is not possible to estimate the sensitive parameters with as much
accuracy as desired. The true values of these parameters may not become known until considerably later when the optimal solution (according to the model) is actually implemented.
Therefore, even after estimating the sensitive parameters as carefully as possible, significant
estimation errors can occur for these parameters along with even larger estimation errors for
the other parameters. This can lead to unfortunate consequences. Perhaps the optimal solution
(according to the model) will not be optimal after all. In fact, it may not even be feasible.
The seriousness of these unfortunate consequences depends somewhat on whether there
is any latitude in the functional constraints in the model. It is useful to make the following
distinction between these constraints.
A soft constraint is a constraint that can be violated a little bit without very serious complications. By contrast, a hard constraint is a constraint that must be satisfied.
Robust optimization is especially designed for dealing with problems with hard constraints.
180 Chapter Five What-If Analysis for Linear Programming
The goal of robust optimization is to find a solution for the model that is guaranteed to remain
feasible and near optimal for all plausible combinations of the actual values for the parameters.
Such a solution is called a robust solution.
This is a daunting goal, but an elaborate theory of robust optimization now has been developed. Much of this theory is beyond the scope of this book, but we will introduce the basic
concept by considering the following straightforward case of independent parameters.
Robust Optimization with Independent Parameters
This case makes four basic assumptions, where assumption 3 defines independent parameters.
1. Each parameter has a range of uncertainty surrounding its estimated value.
2. This parameter can take any value between the minimum and maximum specified by this
range of uncertainty.
3. This value is uninfluenced by the values taken on by the other parameters.
4. All the functional constraints are in either ≤ or ≥ form.
Generally, in robust optimization, each uncertain
parameter should be
assigned its most conservative value within the range
of uncertainty.
To guarantee that the solution will remain feasible regardless of the values taken on by
these parameters within their ranges of uncertainty, we just need to assign the most conservative value to each uncertain parameter and then find the optimal solution for the revised
problem.
Example
We now will illustrate the robust optimization approach for this case of independent parameters by addressing the fifth what-if question posed by Wyndor management.
Question 5: What happens if the production rates of doors and windows at plant 3 are
uncertain?
In particular, because of the brand new production processes used to assemble the windows
and doors at plant 3, management is unsure exactly how many hours will be required to produce each window and each door. The initial estimates (as given in Table 2.1) of 3 hours per
door and 2 hours per window could be off by as much as half an hour each.
In this case, we are dealing with the following two uncertain parameters:
HD3 = production hours required per door in plant 3
HW3 = production hours required per window in plant 3
Because the production processes are different for doors and windows, the value of each
of these parameters is uninfluenced by the value of the other one, so these are independent
parameters. Since the initial estimates of HD3 = 3 and HW3 = 2 could each be off by as much
as half an hour, their ranges of uncertainty are
Range of uncertainty for HD3 = 2.5–3.5 hours
Range of uncertainty for HW3 = 1.5–2.5 hours
By setting all of the uncertain parameters at their
most conservative values
and running Solver again
for the revised problem, a
robust solution has been
found that is guaranteed to
be feasible for all plausible
combinations of actual
values for the uncertain
parameters.
The actual values of these parameters will not become known until some time after production has begun, but a decision is needed now regarding the initial production schedule for the
doors and windows in the three plants.
Recall that the model for the Wyndor problem includes a constraint that the total number
of hours devoted to doors and windows in plant 3 cannot exceed 18 hours per week. This is a
hard constraint (it must be satisfied), as are the corresponding constraints for plants 1 and 2.
Therefore, management would like to choose a production rate for windows and doors that is
guaranteed to be remain feasible and also should be at least nearly optimal.
Applying the procedure for robust optimization with independent parameters, each uncertain parameter is assigned the most conservative value in its range of uncertainty. It is most
conservative to assume each door requires the maximum value in its range of uncertainty,
or 3.5 hours. Similarly, it is most conservative to assume each window requires 2.5 hours.
5.7 Robust Optimization 181
FIGURE 5.26
A
Applying robust optimization to the Wyndor
problem, each uncertain
parameter has been set
to its most conservative value in its range of
uncertainty: HD3 = 3.5
and HW3 = 2.5. Running
Solver then yields this
robust solution.
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
0.857
≤
4
8
Plant 2
0
2
12
≤
12
9
Plant 3
3.5
2.5
18
≤
18
Doors
Windows
Total Profit
0.857
6
$3,257
10
11
12
Units Produced
Making this change in the spreadsheet model and running Solver yields the spreadsheet shown
in Figure 5.26. Producing 0.857 doors and 6 windows per week yields a profit of $3,257 and
is guaranteed to be feasible if the hours required per door and window fall anywhere within
their ranges of uncertainty.
The General Procedure for Robust Optimization
with Independent Parameters
The above example considers a case whether the only uncertain parameters were the coefficients in a single functional constraint. However, robust optimization with independent parameters can also be implemented when there is a range of uncertainty for various other parameters
in the model, including objective coefficients and the right-hand-side of the constraints. Simply
choose the most conservative value for each parameter as follows:
• For each functional constraint in ≤ form, use the maximum value for each uncertain
­coefficient on the left of the ≤ and the minimum value for an uncertain right-hand-side.
• For each functional constraint in ≥ form, use the minimum value for each uncertain
­coefficient on the left of the ≥ and the maximum value for an uncertain right-hand-side.
• For an objective function in maximization form, use the minimum value of each uncertain
coefficient.
• For an objective function in minimization form, use the maximum value for each uncertain
coefficient.
Then run Solver again for the revised model. The resulting solution is guaranteed to be feasible so long as all parameters fall somewhere within their ranges of uncertainty.
Review
Questions
1.
2.
3.
4.
5.
What is meant by a soft constraint?
What is meant by a hard constraint?
What is the goal of robust optimization?
What are the four basic assumptions of robust optimization with independent parameters?
In the general procedure for robust optimization with independent parameters, how should
the value of an uncertain coefficient on the left-hand-side of a functional constraint be chosen
from within its range of uncertainty?
6. In the general procedure for robust optimization with independent parameters, how should
the value of an uncertain coefficient in the objective function be chosen from within its range
of uncertainty?
7. In the general procedure for robust optimization with independent parameters, how should
the value of an uncertain right-hand-side of a functional constraint be chosen from within its
range of uncertainty?
182 Chapter Five What-If Analysis for Linear Programming
5.8
CHANCE CONSTRAINTS WITH ANALYTIC SOLVER
Two shortcomings of robust
optimization are that (1) it
can be extremely conservative, and (2) it can be difficult to accurately identify
an upper and lower bound
for an uncertain parameter.
Chance constraints are useful when dealing with soft
constraints.
The parameters of a linear programming model often remain uncertain until the actual values
of these parameters can be observed at some later time when the adopted solution is implemented for the first time. The preceding section described how robust optimization deals with
this uncertainty by revising the values of the parameters in the model to ensure that the resulting solution will be feasible when it finally is implemented. When dealing with independent
parameters, this involves identifying an upper and lower bound on the possible value of each
uncertain parameter. The estimated value of the parameter then is replaced by whichever of
these two bounds is the more conservative one.
This is a useful approach when dealing with hard constraints, i.e., those constraints that
must be satisfied. However, it does have certain shortcomings. One is that it can be extremely
conservative in tightening the model far more than is realistically necessary. This is especially
true when dealing with large models with hundreds or thousands (perhaps even millions) of
parameters. Another is that it might not be possible to accurately identify an upper and lower
bound for an uncertain parameter. In fact, it might not even have an upper and lower bound.
This is the case, for example, when the underlying probability distribution for a parameter is a
normal distribution, which has long tails with no bounds.
Chance constraints are useful when dealing with soft constraints, i.e., those constraints that
can be violated a little bit without very serious complications. Therefore, rather than requiring
that a soft constraint must be satisfied, it would make sense to require only that it probably
will be satisfied but there is a small chance that it will be violated a little bit. This is what a
chance constraint does.
A chance constraint only requires that the original constraint will be satisfied with some very
high probability while leaving a small chance that it will be violated a little bit.
The Analytic Solver includes the ability to define uncertain parameters. It also can include
chance constraints.
To implement chance constraints in Analytic Solver, each of the uncertain parameters must
first be specified as following a particular probability distribution. This concept will be covered in much more detail in Chapter 13 (Computer Simulation with Analytic Solver). For
the purpose of demonstrating chance constraints now, a simple formula for specifying that a
parameter should follow the uniform distribution (and later the normal distribution) will be
shown. Chapter 13 will consider various more advanced probability distributions and other
methods of defining them using the menus on the Analytic Solver ribbon.
Chance Constraints with the Uniform Distribution
We now will illustrate chance constraints by again addressing the fifth what-if question posed
by Wyndor management in the preceding section.
Question 5: What happens if the production rates of doors and windows at plant 3 are uncertain?
As discussed in the preceding section, the introduction of brand new production processes
for assembling the windows and doors in plant 3 results in uncertainty about how many hours
will be required to assemble each window and each door there. The range of uncertainty for
the hours required per door is 2.5–3.5, while the range of uncertainty for the hours required
per window is 1.5–2.5. If we assume that any value within these ranges is equally likely, this
would mean that each of these parameters follows the uniform distribution.
Analytic Solver includes a function called PsiUniform(Min, Max) to specify that a
value should follow the uniform distribution over the range from Min to Max. As shown in
­Figure 5.27, =PsiUniform(2.5, 3.5) is entered in cell C9 and =PsiUniform(1.5, 2.5) is entered
in cell D9. Analytic Solver then generates a random value from the specified uniform distribution (shown as 2.737 and 2.317 in cells C9 and D9, respectively).
In the original Wyndor problem, it was optimal to produce 2 doors per week and 6 windows
per week. However, with the uncertainty in the production rates at plant 3, this solution could
lead to using too many hours at plant 3. In fact, this happens for the particular random values
that Analytic Solver has generated in cells C9 and D9 of Figure 5.27. At these production
5.8 Chance Constraints with Analytic Solver 183
FIGURE 5.27
A spreadsheet model
for applying chance
constraints to the Wyndor
problem before finding a
final solution. The hours
used per door and per
window produced have
been specified to follow
the uniform distribution
in cells C9 and D9,
respectively.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
2
≤
4
8
Plant 2
0
2
12
≤
12
9
Plant 3
2.737
2.317
19.375
≤
18
Doors
Windows
Total Profit
2
6
$3,600
10
11
12
Units Produced
9
C
D
=PsiUniform(2.5,3.5)
=PsiUniform(1.5,2.5)
rates, 19.375 hours would be used at plant 3 while only 18 hours are available. There still is
a little more work that needs to be done before finding a final solution. Wyndor management
still needs to define a chance constraint for plant 3. Then Analytic Solver will be used to solve
the model to find a solution such that the original constraint would usually be satisfied.
Robust optimization was used in the preceding section to find a solution that guarantees
a feasible solution, with no more than 18 hours used per week in plant 3. However, management now realizes there may be some flexibility with this constraint. For example, there may
actually be some flexibility with the number of hours available for doors and windows in
plant 3. If more than 18 hours were required in some week, overtime could be used to meet
the requirements. However, overtime is expensive and management would like to avoid this
most of the time.
Therefore, in contrast to what was done with robust optimization in the preceding section,
management will no longer absolutely require that this constraint be met. Instead, they have
decided that they would like to require that this constraint be satisfied at least 95 percent of
the time. This can be done using a chance constraint.
To implement this chance constraint in Analytic Solver, choose cell E9 (hours used in
plant 3 in Figure 5.27), and then select <= from Constraints > Chance Constraints on the
Analytic Solver ribbon. This brings up the dialog box shown in Figure 5.28. Cell E9 is already
chosen as the left-hand side of the constraint. Choose cell G9 for the right-hand side of the
constraint. Finally, under Chance, choose the probability that the constraint should be satisfied.
The default value of 0.95 is entered in Figure 5.28. Click OK to accept this chance constraint.
An important option for chance constraints in Analytic Solver is found under the Platform
tab in the Model pane. Auto Adjust Chance Constraints, as shown in Figure 5.29, can be set
to either True or False (and is False by default). Set at False, the solution will often be very
conservative in the sense that it usually will give a probability that actually is considerably
FIGURE 5.28
This Add Constraint
dialog box specifies a
chance constraint that
E9 is required to be less
than or equal to G9 with
probability 0.95.
184 Chapter Five What-If Analysis for Linear Programming
FIGURE 5.29
The Platform tab of
Analytic Solver’s Model
pane that shows the Auto
Adjust Chance Constraints
option set to True.
Setting the Auto Adjust
Chance Constraints option
to True results in a better
solution while still meeting the requirements of the
chance constraints.
Generally, robust optimization yields a more conservative solution than when
using chance constraints
with uniform distributions.
higher than the one specified in the chance box shown in Figure 5.28. Setting this option to
True causes Solver to automatically re-optimize the initial conservative solution to find a better solution that still meets the requirements of each chance constraint with the probability
specified in the chance box. Therefore, this option generally should be set to True to find better (albeit less conservative) solutions.
With Auto Adjust Chance Constraints set to true, the model is then solved by clicking on
the Optimize button on the Analytic Solver ribbon. Analytic Solver will then find a solution
that maximizes TotalProfit (G12) while meeting all the hard constraints (E7:E8 ≤ G7:G8) and
yields at least a 95% chance of satisfying the original constraint (E9 ≤G9). This solution is
shown in Figure 5.30.
It is interesting to compare the solution found by robust optimization and shown in
­Figure 5.26 with the solution shown in Figure 5.30 when a chance constraint has been used.
With robust optimization (requiring that no more than 18 hours ever will be used in any
week), 0.857 doors and 6 windows are produced per week, yielding a profit of $3,257. When
using a chance constraint (requiring just a 95% chance that no more than 18 hours are used in
plant 3 in any given week), 1.103 doors and 6 windows should be produced per week, yielding a higher profit of $3,331. This comparison between the two approaches is typical in other
cases as well. In general, robust optimization yields a very conservative solution while chance
constraints with uniform distributions allow for a better solution if there is some flexibility in
the original constraints.
Chance Constraints with the Normal Distribution
The robust optimization procedure discussed in the preceding section requires that each uncertain parameter have a specified range of uncertainty (a minimum and a maximum). While the
FIGURE 5.30
The spreadsheet obtained
after solving the Wyndor
problem with Analytic
Solver to maximize the
TotalProfit (G12) while
meeting the hard constraints (E7:E8 ≤ G7:G8)
and having at least a 95%
chance of satisfying
the original constraint
(E9 ≤ G9).
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
1.1027
≤
4
8
Plant 2
0
2
12
≤
12
9
Plant 3
3.108
2.193
16.584
≤
18
Doors
Windows
Total Profit
1.103
6
$3,331
10
11
12
Units Produced
5.8 Chance Constraints with Analytic Solver 185
A more conservative solution is necessary to satisfy
a chance constraint when
the variability is increased
in the corresponding
parameter.
Review
Questions
uniform distribution for an uncertain parameter meets the requirement of having a range of
uncertainty, many other distributions do not. For example, the normal distribution has long
tails with no minimum or maximum. Therefore, the procedure for robust optimization discussed in the preceding section cannot be used. However, chance constraints can still be used.
Suppose management is now more uncertain about the hours used per door and window at
plant 3. The uniform distribution assumed a maximum deviation of 0.5 hours from the original
estimate, so the range for this distribution is 2.5–3.5 hours for doors and 1.5–2.5 hours for
windows. Management now feels that these parameters should follow the normal distribution,
with a standard deviation of 0.5 hours. Given the long tails of the normal distribution, this
leads to the possibility that the hours used could deviate well more than 0.5 hours from the
original estimates.
Analytic Solver includes a function called PsiNormal(Mean, Standard Deviation) to specify the normal distribution. Figure 5.31, shows the model revised to use the normal distribution, with =PsiNormal(3, 0.5) entered in cell C9 and =PsiNormal(2, 0.5) entered in cell D9.
Analytic Solver has generated a random value from the specified normal distribution (shown
as 2.543 and 2.256 in cells C9 and D9, respectively). The chance constraint is defined in
the same way as with the uniform distribution, with 0.95 again entered into the chance box.
After clicking the Optimize button on the Analytic Solver ribbon, Figure 5.31 shows the new
optimal solution. Now 0.298 doors and 6 windows should be produced per week, leading to a
profit of $3,089. This leads to a 95% chance of satisfying the original constraint for plant 3.
It is interesting to compare the solution in Figure 5.30 (which assumes uniform distributions with a maximum deviation of 0.5 hours) to the solution in Figure 5.31 (which assumes
normal distributions with a standard deviation of 0.5 hours). With normal distributions, just
0.298 doors are produced per week instead of 1.103 when using uniform distributions, leading
to a smaller profit of $3,089 as compared to $3,331. This more conservative solution is necessary to maintain the 95% probability of satisfying the original constraint for plant 3 given the
higher variability in the production rates with this particular normal distribution. (A normal
distribution with a considerably smaller standard deviation would lead to a considerably less
conservative solution.)
1. Are chance constraints more useful for hard constraints or soft constraints?
2. What Analytic Solver function is used to specify a value that follows the uniform distribution?
3. Which approach will lead to a more conservative solution—robust optimization with specified
ranges of uncertainty or using chance constraints with uniform distributions over these same
ranges of uncertainty?
4. What Analytic Solver function is used to specify a value that follows the normal distribution?
FIGURE 5.31
The spreadsheet obtained
after solving the Wyndor
problem with a chance
constraint for plant 3
when the hours used
per door and per window
produced have been
specified to follow the
normal distribution.
A
1
B
C
D
E
F
G
Wyndor Glass Co. Product-Mix Problem
2
3
4
Unit Profit
Doors
Windows
$300
$500
5
6
Hours Used per Unit Produced
Hours
Hours
Used
Available
7
Plant 1
1
0
0.2982
≤
4
8
Plant 2
0
2
12
≤
12
9
Plant 3
2.543
2.256
14.295
≤
18
Doors
Windows
Total Profit
0.298
6
$3,089
10
11
12
Units Produced
C
9
=PsiNormal(3, 0.5)
D
=PsiNormal(2, 0.5)
186 Chapter Five What-If Analysis for Linear Programming
5.9 Summary
What-if analysis is analysis done after finding an optimal solution for the original version of the basic
model. This analysis provides important insights to help guide managerial decision making. This chapter describes how this is done when the basic model is a linear programming model. The spreadsheet
for the model, the parameter analysis report available with Analytic Solver, and the sensitivity report
generated by Solver all play a central role in this process.
The coefficients in the objective function typically represent quantities that can only be roughly
estimated when the model is formulated. Will the optimal solution obtained from the model be the correct one if the true value of one of these coefficients is significantly different from the estimate used
in the model? The spreadsheet can be used to quickly check specific changes in the coefficient. The
parameter analysis report enables the systematic investigation of many trial values for this coefficient.
For a broader investigation, the allowable range for each coefficient identifies the interval within which
the true value must lie in order for this solution to still be the correct optimal solution. These ranges are
easily calculated from the data in the sensitivity report provided by Solver.
What happens if there are significant inaccuracies in the estimates of two or more coefficients in
the objective function? Specific simultaneous changes can be checked with the spreadsheet. A two-way
parameter analysis report can systematically investigate various simultaneous changes in two coefficients. To go further, the 100 percent rule for simultaneous changes in objective function coefficients
provides a convenient way of checking whole ranges of simultaneous changes, again by using the data
in Solver’s sensitivity report.
What-if analysis usually extends to considering the effect of changes in the functional constraints
as well. Occasionally, changes in the coefficients of these constraints will be considered because of the
uncertainty in their original estimates. More frequently, the changes considered will be in the right-hand
sides of the constraints. The right-hand sides frequently represent managerial policy decisions. In such
cases, shadow prices provide valuable guidance to management about the potential effects of altering
these policy decisions. The shadow price for each constraint is easily found by using the spreadsheet, a
parameter analysis report, or the sensitivity report.
Shadow price analysis can be validly applied to investigate possible changes in right-hand sides as
long as these changes are not too large. The allowable range for each right-hand side indicates just how
far it can be changed, assuming no other changes are made. If other right-hand sides are changed as well,
the 100 percent rule for simultaneous changes in right-hand sides enables checking whether the changes
definitely are not too large. Solver’s sensitivity report provides the key information needed to find each
allowable range or to apply this 100 percent rule. Both the spreadsheet and the parameter analysis report
also can sometimes be used to help investigate these simultaneous changes.
In some cases, there may be so much uncertainty about what the true values of the model parameters
will turn out to be, management may want to be sure that the proposed solution will at least turn out to
be feasible. This is particularly important if it would be completely unacceptable to violate any of the
constraints even a little bit. Robust optimization provides a way of identifying a solution that is virtually guaranteed to be feasible as well as nearly optimal regardless of what the true values of the model
parameters turn out to be.
If the model includes some constraints that actually can be violated a little bit without serious complications, chance constraints can be used to represent these constraints. A chance constraint requires
that the original constraint probably will be satisfied but allows a small probability for violating the
original constraint by a small amount.
Glossary
allowable range for an objective function
coefficient The range of values for a particular
coefficient in the objective function over which
the optimal solution for the original model
remains optimal. (Section 5.3), 160
allowable range for the right-hand side The
range of values for the right-hand side of a
functional constraint over which this constraint’s
shadow price remains valid. (Section 5.5), 172
chance constraint A modification of a
constraint which only requires that the original
constraint will be satisfied with some very high
probability while leaving a small chance that it
will violated a little bit. (Section 5.8), 182
hard constraint A constraint that must be
satisfied. (Section 5.7) 179
independent parameters Parameters whose
values are uninfluenced by the values taken on
by the other parameters. (Section 5.7), 180
parameter cell A data cell containing a
parameter that will be systematically varied
when generating a parameter analysis report.
(Section 5.3), 158
parameters of the model The parameters
of a linear programming model are the constants
(coefficients or right-hand sides) in the functional constraints and the objective function.
(Section 5.1), 152
Chapter 5 Solved Problem 187
range of uncertainty The range between the
minimum and maximum possible value for a
parameter. (Section 5.7), 180
robust optimization An optimization procedure that finds a solution that is guaranteed to
remain feasible and near optimal for all plausible
combinations of the actual values for the
parameters. (Section 5.7) 180
robust solution A solution found by using
robust optimization. (Section 5.7) 180
sensitive parameter A parameter is considered
sensitive if even a small change in its value can
change the optimal solution. (Section 5.1), 152
sensitivity analysis The part of what-if analysis that focuses on individual parameters of the
model. It involves checking how sensitive the
optimal solution is to the value of each parameter.
(Section 5.1), 153
shadow price The shadow price for a functional constraint is the rate at which the optimal
value of the objective function can be increased
by increasing the right-hand side of the constraint
by a small amount. (Section 5.5), 172
soft constraint A constraint that can be violated
a little bit without very serious complications.
(Section 5.7), 179
what-if analysis Analysis that addresses questions about what would happen to the optimal
solution if different assumptions were made about
future conditions. (Chapter introduction), 151
Learning Aids for This Chapter
All learning aids are available at www.mhhe.com/Hillier6e.
Excel Add-in:
Excel Files:
Analytic Solver
Wyndor Example
Supplement to This Chapter:
Profit & Gambit Example
Reduced Costs
Solved Problem
The solution is available at www.mhhe.com/Hillier6e.
5.S1. Sensitivity Analysis at Stickley Furniture
Stickley Furniture is a manufacturer of fine hand-crafted furniture. During the next production period (the eight-hour shift
tomorrow), management is considering producing dining room
tables, dining room chairs, and/or bookcases. The time required
for each item to go through the two stages of production (assembly and finishing), the amount of wood required (fine cherry
wood), and the corresponding unit profits are given in the following table, along with the amount of each resource available
in the upcoming production period.
Tables
Chairs
Bookcases
Available
Assembly
(minutes)
80
40
50
8,100
Finishing)
(minutes)
30
20
30
4,500
Wood
(pounds)
80
10
50
9,000
$360
$125
$300
Unit profit
a. Suppose the profit per table increases by $100. Will this
change the optimal production quantities? What can be said
about the change in total profit?
b. Suppose the profit per chair increases by $100. Will this
change the optimal production quantities? What can be said
about the change in total profit?
c. Suppose the profit per table increases by $90 and the profit
per bookcase decreases by $50. Will this change the optimal
production quantities? What can be said about the change in
total profit?
d. Suppose a worker in the assembly department calls in sick, so
eight fewer hours will be available tomorrow in the assembly
department. How much would this affect total profit? Would
it change the optimal production quantities?
e. Explain why the shadow price for the wood constraint is zero.
After formulating a linear programming model to determine
the production levels that would maximize profit, the solved
model and the corresponding sensitivity report are shown below.
f. A new worker has been hired who is trained to do both assembly and finishing. She will split her time between the two
areas, so there now are four additional hours available tomorrow in both assembly and finishing. How much would this
affect total profit? Would this change the optimal production
quantities?
g. Based on the sensitivity report given below, is it wise to have
the new worker in part f split her time equally between assembly and finishing, or would some other plan be better?
188 Chapter Five What-If Analysis for Linear Programming
B
3
4
Unit Profit
C
D
E
Tables
Chairs
Bookcases
F
$360
$125
$300
G
H
5
Resources Required per Unit
6
Used
Available
7
Assembly (minutes)
80
40
50
8,100
8,100
8
Finishing (minutes)
30
20
30
4,500
4,500
9
Woods (pounds)
80
10
50
8,100
9,000
Tables
Chairs
Bookcases
Total Profit
20
0
130
$46,200
Final
Value
Reduced
Cost
10
11
Production
12
Variable Cells
Cell
Name
$C$12
Production Tables
20
$D$12
Production Chairs
0
$E$12
Production Bookcases
Objective
Coefficient
0
360
−88.333
130
Allowable
Increase
0
Allowable
Decrease
120
60
125
88.333
300
60
1E + 30
75
Constraints
Final
Value
Shadow
Price
Constraint
R. H. Side
Cell
Name
$F$7
Assembly (minutes) Used
8,100
2
8,100
$F$8
Finishing (minutes) Used
4,500
6.67
$F$9
Wood (pounds) Used
8,100
0
h. Use Analytic Solver and a parameter analysis report to determine how the optimal production quantities and total
profit will change depending on how the new worker in
part f allocates her time between assembly and finishing.
Allowable
Increase
Allowable
Decrease
900
600
4,500
360
1,462.5
9,000
1E + 30
900
In particular, assume 0, 1, 2, . . . , or 8 hours are added to
assembly, with a corresponding 8, 7, 6, . . . , or 0 hours
added to finishing. (The original spreadsheet is contained
in www.mhhe.com/Hillier6e.)
Problems
We have inserted the symbol AS to the left of each problem (or
its parts) where Analytic Solver is required. An asterisk on the
problem number indicates that at least a partial answer is given
in the back of the book.
5.1.*
One of the products of the G. A. Tanner Company is a
special kind of toy that provides an estimated unit profit of $3.
Because of a large demand for this toy, management would like
to increase its production rate from the current level of 1,000
per day. However, a limited supply of two subassemblies (A
and B) from vendors makes this difficult. Each toy requires two
subassemblies of type A, but the vendor providing these subassemblies would only be able to increase its supply rate from the
current 2,000 per day to a maximum of 3,000 per day. Each toy
requires only one subassembly of type B, but the vendor providing these subassemblies would be unable to increase its supply
rate above the current level of 1,000 per day.
Because no other vendors currently are available to provide
these subassemblies, management is considering initiating a new
production process internally that would simultaneously produce
an equal number of subassemblies of the two types to supplement
the supply from the two vendors. It is estimated that the company’s
cost for producing one subassembly of each type would be $2.50
more than the cost of purchasing these subassemblies from the two
vendors. Management wants to determine both the production rate
of the toy and the production rate of each pair of subassemblies
(one A and one B) that would maximize the total profit.
Viewing this problem as a resource-allocation problem, one
of the company’s managers has organized its data as follows:
Resource Usage per
Unit of Each Activity
Resource
Produce
Toys
Produce
Subassemblies
Amount of
Resource
Available
Subassembly A
2
−1
3,000
Subassembly B
1
−1
1,000
Unit profit
$3
−$2.50
Chapter 5 Problems 189
a. Formulate and solve a spreadsheet model for this
problem.
b. Since the stated unit profits for the two activities
are only estimates, management wants to know how
much each of these estimates can be off before the
optimal solution would change. Begin exploring this
question for the first activity (producing toys) by
using the spreadsheet and Solver to manually generate a table that gives the optimal solution and total
profit as the unit profit for this activity increases in
50¢ increments from $2.00 to $4.00. What conclusion can be drawn about how much the estimate of
this unit profit can differ in each direction from its
original value of $3.00 before the optimal solution
would change?
c. Repeat part b for the second activity (producing
subassemblies) by generating a table as the unit
profit for this activity increases in 50¢ increments
from –$3.50 to –$1.50 (with the unit profit for the
first activity fixed at $3).
AS
d. Use a parameter analysis report to systematically
generate all the data requested in parts b and c,
except use 25¢ increments instead of 50¢ increments. Use these data to refine your conclusions in
parts b and c.
e. Use Solver’s sensitivity report to find the allowable
range for the unit profit of each activity.
AS
f. Use a two-way parameter analysis report to systematically generate the total profit as the unit profits
of the two activities are changed simultaneously as
described in parts b and c.
g. Use the information provided by Solver’s sensitivity report to describe how far the unit profits of the
two activities can change simultaneously before the
optimal solution might change.
5.2.
Consider a resource-allocation problem having the following data:
Resource Usage per
Unit of Each Activity
Resource
1
2
Amount of
Resource
Available
1
1
2
10
1
3
12
$2
$5
2
Unit profit
The objective is to determine the number of units of each
activity to undertake so as to maximize the total profit.
While doing what-if analysis, you learn that the estimates of
the unit profits are accurate only to within ± 50 percent. In other
words, the ranges of likely values for these unit profits are $1 to
$3 for activity 1 and $2.50 to $7.50 for activity 2.
a. Formulate a spreadsheet model for this problem
based on the original estimates of the unit profits.
Then use Solver to find an optimal solution and to
generate the sensitivity report.
b. Use the spreadsheet and Solver to check whether
this optimal solution remains optimal if the unit
profit for activity 1 changes from $2 to $1. From $2
to $3.
c. Also check whether the optimal solution remains
optimal if the unit profit for activity 1 still is $2
but the unit profit for activity 2 changes from $5 to
$2.50. From $5 to $7.50.
AS
d. Use a parameter analysis report to systematically
generate the optimal solution and total profit as the
unit profit of activity 1 increases in 20 cent increments from $1 to $3 (without changing the unit
profit of activity 2). Then do the same as the unit
profit of activity 2 increases in 50 cent increments
from $2.50 to $7.50 (without changing the unit
profit of activity 1). Use these results to estimate the
allowable range for the unit profit of each activity.
e. Use the sensitivity report to find the allowable
range for the unit profit of each activity. Then use
these ranges to check your results in parts b-d.
AS
f. Use a two-way parameter analysis report to systematically generate the optimal total profit as the unit
profit of the two activities are changes simultaneously as described in part d.
5.3.
Consider the Big M Co. problem presented in Section
3.5, including the spreadsheet in Figure 3.10 showing its formulation and optimal solution.
There is some uncertainty about what the unit costs will
be for shipping through the various shipping lanes. Therefore,
before adopting the optimal solution in Figure 3.10, management wants additional information about the effect of inaccuracies in estimating these unit costs.
Use Solver to generate the sensitivity report preparatory to
addressing the following questions.
a. Which of the unit shipping costs given in Table 3.9
has the smallest margin for error without invalidating the optimal solution given in Figure 3.10?
Where should the greatest effort be placed in estimating the unit shipping costs?
b. What is the allowable range for each of the unit
shipping costs?
c. How should the allowable range be interpreted to
management?
d. If the estimates change for more than one of the
unit shipping costs, how can you use the sensitivity report to determine whether the optimal solution
might change?
5.4.*
Consider the Union Airways problem presented in Section 3.3, including the spreadsheet in Figure 3.5 showing its formulation and optimal solution.
Management is about to begin negotiations on a new contract
with the union that represents the company’s customer service
agents. This might result in some small changes in the daily
costs per agent given in Table 3.5 for the various shifts. Several
possible changes listed below are being considered separately. In
each case, management would like to know whether the change
might result in the solution in Figure 3.5 no longer being optimal. Answer this question in parts a to e by using the spreadsheet and Solver directly. If the optimal solution changes, record
the new solution.
a. The daily cost per agent for shift 2 changes from
$160 to $165.
190 Chapter Five What-If Analysis for Linear Programming
b. The daily cost per agent for shift 4 changes from
$180 to $170.
c. The changes in parts a and b both occur.
d. The daily cost per agent increases by $4 for shifts 2,
4, and 5, but decreases by $4 for shifts 1 and 3.
e. The daily cost per agent increases by 2 percent for
each shift.
f. Use Solver to generate the sensitivity report for this
problem. Suppose that the above changes are being
considered later without having the spreadsheet
model immediately available on a computer. Show
in each case how the sensitivity report can be used
to check whether the original optimal solution must
still be optimal.
AS
g. For each of the five shifts in turn, use a parameter
analysis report to systematically generate the optimal solution and total cost when the only change is
that the daily cost per agent on that shift increases in
$3 increments from $15 less than the current cost up
to $15 more than the current cost.
5.5.
Consider the Think-Big Development Co. problem presented in Section 3.2, including the spreadsheet in Figure 3.3
showing its formulation and optimal solution. In parts a-g, use
the spreadsheet and Solver to check whether the optimal solution
would change and, if so, what the new optimal solution would
be, if the estimates in Table 3.3 of the net present values of the
projects were to be changed in each of the following ways. (Consider each part by itself.)
a. The net present value of project 1 (a high-rise office
building) increases by $200,000.
b. The net present value of project 2 (a hotel) increases
by $200,000.
c. The net present value of project 1 decreases by
$5 million.
d. The net present value of project 3 (a shopping center) decreases by $200,000.
e. All three changes in parts b, c, and d occur
simultaneously.
f. The net present values of projects 1, 2, and 3 change
to $46 million, $69 million, and $49 million,
respectively.
g. The net present values of projects 1, 2, and 3 change
to $54 million, $84 million, and $60 million,
respectively.
A
1
2
3
4
5
6
7
8
9
10
Unit Profit
Molding (minutes)
Finishing (minutes)
Clay (ounces)
Production
B
Plates
$3.10
4
8
5
h. Use Solver to generate the sensitivity report for
this problem. For each of the preceding parts, suppose that the change occurs later without having the
spreadsheet model immediately available on a computer. Show in each case how the sensitivity report
can be used to check whether the original optimal
solution must still be optimal.
AS
i. For each of the three projects in turn, use a parameter analysis report to systematically generate the
optimal solution and the total net present value
when the only change is that the net present value of
that project increases in $1 million increments from
$5 million less than the current value up to $5 million more than the current value.
5.6.
University Ceramics manufactures plates, mugs, and
steins that include the campus name and logo for sale in campus bookstores. The time required for each item to go through
the two stages of production (molding and finishing), the material required (clay), and the corresponding unit profits are given
in the following table, along with the amount of each resource
available for tomorrow’s eight-hour shift.
Plates
Mugs
Steins
Available
Molding (minutes)
4
6
3
2,400
Finishing (minutes)
8
14
12
7,200
Clay (ounces)
5
4
3
3,000
$3.10
$4.75
$4.00
Unit Profit
A linear programming model has been formulated in a
spreadsheet to determine the production levels for tomorrow that
would maximize profit. The solved spreadsheet model and corresponding sensitivity report are shown below.
For each of the following parts, answer the question as specifically and completely as is possible without re-solving the
problem with Solver. Note: Each part is independent (i.e., any
change made in one part does not apply to any other parts).
a. Suppose the profit per plate decreases from $3.10
to $2.80. Will this change the optimal production
quantities? What can be said about the change in
total profit?
b. Suppose the profit per stein increases by $0.30 and
the profit per plate decreases by $0.25. Will this
change the optimal production quantities? What can
be said about the change in total profit?
C
Mugs
$4.75
Resource Required per Unit
6
14
4
Plates
300
Mugs
0
D
Steins
$4.00
3
12
3
Steins
400
E
Used
2,400
7,200
2,700
F
G
<=
<=
<=
Available
2,400
7,200
3,000
Total Profit
$2,530
Chapter 5 Problems 191
Variable Cells
Cell
Final
Value
Name
$B$10
Production Plates
300
$C$10
Production Mugs
0
$D$10
Production Steins
400
Reduced
Cost
Objective
Coefficient
Allowable
Increase
Allowable
Decrease
3.10
2.23
0.37
4.75
0.46
4.00
0.65
1.37
Allowable
Increase
Allowable
Decrease
0
−0.46
0
Constraints
Cell
Name
Final
Value
Shadow
Price
$E$5
Molding (minutes) Used
2,400
0.22
2400
200
600
$E$6
Finishing (minutes) Used
7,200
0.28
7200
2400
2400
$E$7
Clay (ounces) Used
2,700
0
3000
1E + 30
c. Suppose a worker in the molding department calls
in sick. Now eight fewer hours are available tomorrow in the molding department. How much would
this affect total profit? Would it change the optimal
production quantities?
d. Suppose one of the workers in the molding department is also trained to do finishing. Would it be a
good idea to have this worker shift some of her time
from the molding department to the finishing department? Indicate the rate at which this would increase
or decrease total profit per minute shifted. How many
minutes can be shifted before this rate might change?
e. The allowable decrease for the mugs’ objective
coefficient and for the available clay constraint are
both missing from the sensitivity report. What numbers should be there? Explain how you were able to
deduce each number.
5.7.
Ken and Larry, Inc., supplies its ice cream parlors with
three flavors of ice cream: chocolate, vanilla, and banana. Due
to extremely hot weather and a high demand for its products, the
company has run short of its supply of ingredients: milk, sugar,
and cream. Hence, they will not be able to fill all the orders
received from their retail outlets, the ice cream parlors. Due to
these circumstances, the company has decided to choose the
amount of each flavor to produce that will maximize total profit,
given the constraints on the supply of the basic ingredients.
The chocolate, vanilla, and banana flavors generate, respectively, $1.00, $0.90, and $0.95 of profit per gallon sold. The
A
1
2
Unit Profit
Constraint
R. H. Side
company has only 200 gallons of milk, 150 pounds of sugar, and
60 gallons of cream left in its inventory. The linear programming
formulation for this problem is shown below in algebraic form.
Let
C = Gallons of chocolate ice cream produced
V = Gallons of vanilla ice cream produced
​​
​​ ​ ​​
B = Gallons of banana ice cream produced
Maximize
Profit = 1.00C + 0.90V + 0.95B
subject to
Milk: 0.45 C + 0.50 V + 0.40 B ≤ 200 gallons
​​Sugar: 0.50 C​ ​  ​  +​ ​  0.40 V​ ​  +​ ​  0.40 B​ ​  ≤​ ​  150 pounds​
​​​
Cream: 0.10 C + 0.15 V + 0.20 B ≤ 60 gallons
and
​​C ≥ 0​
V ≥ 0​
B ≥ 0​​
This problem was solved using Solver. The spreadsheet
(already solved) and the sensitivity report are shown below.
(Note: The numbers in the sensitivity report for the milk constraint are missing on purpose, since you will be asked to fill in
these numbers in part f.)
For each of the following parts, answer the question as specifically and completely as possible without solving the problem again with Solver. Note: Each part is independent (i.e., any
change made to the model in one part does not apply to any other
parts).
B
C
D
Chocolate
Vanilla
Banana
$1.00
$0.90
$0.95
E
F
G
3
4
Resource
Resources Used per Gallon Produced
Used
Available
5
Milk
0.45
0.5
0.4
180
200
6
Sugar
0.5
0.4
0.4
150
150
7
Cream
0.1
0.15
0.2
60
60
Chocolate
Vanilla
Banana
Total Profit
0
300
75
$341.25
8
9
10
Gallons Produced
192 Chapter Five What-If Analysis for Linear Programming
Variable Cells
Cell
Final
Value
Name
Reduced
Cost
Objective
Coefficient
Allowable
Increase
Allowable
Decrease
−0.0375
1
0.0375
1E + 30
$B$10
Gallons Produced Chocolate
0
$C$10
Gallons Produced Vanilla
300
0
0.9
0.05
0.0125
$D$10
Gallons Produced Banana
75
0
0.95
0.0214
0.05
Constraints
Cell
Name
Final
Value
Shadow
Price
Constraint
R. H. Side
Allowable
Increase
1.875
150
10
30
60
15
3.75
$E$5
Milk Used
$E$6
Sugar Used
150
$E$7
Cream Used
60
1
a. What is the optimal solution and total profit?
b. Suppose the profit per gallon of banana changes to
$1.00. Will the optimal solution change and what
can be said about the effect on total profit?
c. Suppose the profit per gallon of banana changes to
92¢. Will the optimal solution change and what can
be said about the effect on total profit?
d. Suppose the company discovers that three gallons
of cream have gone sour and so must be thrown out.
Will the optimal solution change and what can be
said about the effect on total profit?
e. Suppose the company has the opportunity to buy an
additional 15 pounds of sugar at a total cost of $15.
Should it do so? Explain.
f. Fill in all the sensitivity report information for the
milk constraint, given just the optimal solution for
the problem. Explain how you were able to deduce
each number.
5.8.
Colonial Furniture produces hand-crafted colonial style
furniture. Plans are now being made for the production of rocking chairs, dining room tables, and/or armoires over the next week.
These products go through two stages of production (assembly and
finishing). The table in the next column gives the time required
for each item to go through these two stages, the amount of wood
required (fine cherry wood), and the corresponding unit profits,
along with the amount of each resource available next week.
A
1
2
3
4
5
6
7
8
9
10
11
12
Unit Profit
Assembly (minutes)
Finishing (minutes)
Wood (pounds)
Production
B
Rocking
Chair
$240
Rocking
Chair
Dining
Room Table
Armoire Available
Assembly
(minutes)
100
180
120
3,600
Finishing
(minutes)
60
80
80
2,000
Wood
(pounds)
30
180
120
4,000
$240
$720
$600
Unit Profit
A linear programming model has been formulated in a
spreadsheet to determine the production levels that would maximize profit. The solved spreadsheet model and corresponding
sensitivity report are shown below.
For each of the following parts, answer the question as specifically and completely as is possible without re-solving the problem with Solver. Note: Each part is independent (i.e., any change
made in one problem part does not apply to any other parts).
a. Suppose the profit per armoire decreases by $50.
Will this change the optimal production quantities?
What can be said about the change in total profit?
b. Suppose the profit per table decreases by $60 and
the profit per armoire increases by $90. Will this
change the optimal production quantities? What can
be said about the change in total profit?
C
Dining Room
Table
$720
Resource Required per Unit
100
180
60
80
30
180
Rocking
Chair
0
Allowable
Decrease
Dining Room
Table
10
D
E
F
G
<=
<=
<=
Available
3,600
2,000
4,000
Armoire
$600
120
80
120
Armoire
15
Used
3,600
2,000
3,600
Total Profit
$16,200
Chapter 5 Problems 193
Variable Cells
Cell
Name
Final
Value
Reduced
Cost
Objective
Coefficient
Allowable
Increase
Allowable
Decrease
$B$12
Production Chair
0
−230
240
230
1E + 30
$C$12
Production Table
10
0
720
180
120
$D$12
Production Armoire
15
0
600
120
120
Constraints
Cell
Name
Final
Value
Shadow
Price
Constraint
R. H. Side
Allowable
Increase
Allowable
Decrease
$E$6
Assembly (minutes) Used
3,600
2.00
3,600
400
600
$E$7
Finishing (minutes) Used
2,000
4.50
2,000
400
400
$E$8
Wood (pounds) Used
3,600
c. Suppose a part-time worker in the assembly department calls in sick for one day, so that now four fewer
hours are available that day in the assembly department. How much would this affect total profit?
Would it change the optimal production quantities?
d. Suppose one of the workers in the assembly department is also trained to do finishing. Would it be
a good idea to have this worker shift some of his
time from the assembly department to the finishing
department? Indicate the rate at which this would
increase or decrease total profit per minute shifted.
How many minutes can be shifted before this rate
might change?
e. The shadow price and allowable range for the wood
constraint are missing from the sensitivity report.
What numbers should be there? Explain how you
were able to deduce each number.
5.9.
David, LaDeana, and Lydia are the sole partners and
workers in a company that produces fine clocks. David and
LaDeana are each available to work a maximum of 40 hours per
week at the company, while Lydia is available to work a maximum of 20 hours per week.
The company makes two different types of clocks: a grandfather clock and a wall clock. To make a clock, David (a mechanical engineer) assembles the inside mechanical parts of the clock
while LaDeana (a woodworker) produces the hand-carved wood
casings. Lydia is responsible for taking orders and shipping the
clocks. The amount of time required for each of these tasks is
shown next.
Time Required
Task
Grandfather Clock Wall Clock
Assemble clock mechanism
6 hours
4 hours
Carve wood casing
8 hours
4 hours
Shipping
3 hours
3 hours
Each grandfather clock built and shipped yields a profit of
$300, while each wall clock yields a profit of $200.
4,000
The three partners now want to determine how many clocks
of each type should be produced per week to maximize the total
profit.
a. Formulate a linear programming model in algebraic
form for this problem.
b. Use graphical analysis to solve the model. Then
check if the optimal solution would change if the
unit profit for grandfather clocks were changed from
$300 to $375 (with no other changes in the model).
Then check if the optimal solution would change if,
in addition to this change in the unit profit for the
grandfather clocks, the estimated unit profit for wall
clocks also changed from $200 to $175.
c. Formulate and solve the original version of this
model on a spreadsheet.
d. Use Solver to check the effect of the changes specified in part b.
AS
e. Use a parameter analysis report to systematically
generate the optimal solution and total profit as the
unit profit for grandfather clocks is increased in $20
increments from $150 to $450 (with no change in
the unit profit for wall clocks). Then do the same
as the unit profit for wall clocks is increased in $20
increments from $50 to $350 (with no change in the
unit profit for grandfather clocks). Use this information to estimate the allowable range for the unit
profit of each type of clock.
AS
f. Use a two-way parameter analysis report to systematically generate the optimal total profit as the
unit profits for the two types of clocks are changed
simultaneously as specified in part e, except use $50
increments instead of $20 increments.
g. For each of the three partners in turn, use Solver to
determine the effect on the optimal solution and the
total profit if that partner alone were to increase his
or her maximum number of work hours available
per week by 5 hours.
AS
h. Use a parameter analysis report to systematically
generate the optimal solution and the total profit
194 Chapter Five What-If Analysis for Linear Programming
when the only change is that David’s maximum
number of hours available to work per week changes
to each of the following values: 35, 37, 39, 41, 43,
45. Then do the same when the only change is that
LaDeana’s maximum number of hours available to
work per week changes in the same way. Then do
the same when the only change is that Lydia’s maximum number of hours available to work per week
changes to each of the following values: 15, 17, 19,
21, 23, 25.
i. Generate a sensitivity report and use it to determine
the allowable range for the unit profit for each type
of clock and the allowable range for the maximum
number of hours each partner is available to work
per week.
j. To increase the total profit, the three partners have
agreed that one of them will slightly increase the
maximum number of hours available to work per
week. The choice of which one will be based on
which one would increase the total profit the most.
Use the sensitivity report to make this choice.
(Assume no change in the original estimates of the
unit profits.)
k. Explain why one of the shadow prices is equal to
zero.
l. Can the shadow prices in the sensitivity report be
validly used to determine the effect if Lydia were to
change her maximum number of hours available to
work per week from 20 to 25? If so, what would be
the increase in the total profit?
m. Repeat part l if, in addition to the change for Lydia,
David also were to change his maximum number of
hours available to work per week from 40 to 35.
n. Use graphical analysis to verify your answer in
part m.
5.10.* Reconsider Problem 5.1. After further negotiations with
each vendor, management of the G. A. Tanner Company has
learned that either of them would be willing to consider increasing their supply of their respective subassemblies over the previously stated maxima (3,000 subassemblies of type A per day
and 1,000 of type B per day) if the company would pay a small
premium over the regular price for the extra subassemblies. The
size of the premium for each type of subassembly remains to
be negotiated. The demand for the toy being produced is sufficiently high that 2,500 per day could be sold if the supply of
subassemblies could be increased enough to support this production rate. Assume that the original estimates of unit profits given
in Problem 5.1 are accurate.
a. Formulate and solve a spreadsheet model for this
problem with the original maximum supply levels and the additional constraint that no more than
2,500 toys should be produced per day.
b. Without considering the premium, use the spreadsheet and Solver to determine the shadow price for
the subassembly A constraint by solving the model
again after increasing the maximum supply by one.
Use this shadow price to determine the maximum
premium that the company should be willing to pay
for each subassembly of this type.
c. Repeat part b for the subassembly B constraint.
d. Estimate how much the maximum supply of subassemblies of type A could be increased before the
shadow price (and the corresponding premium)
found in part b would no longer be valid by using
a parameter analysis report to generate the optimal
solution and total profit (excluding the premium) as
the maximum supply increases in increments of 100
from 3,000 to 4,000.
AS
e. Repeat part d for subassemblies of type B by using
a parameter analysis report as the maximum supply
increases in increments of 100 from 1,000 to 2,000.
f. Use Solver’s sensitivity report to determine the
shadow price for each of the subassembly constraints
and the allowable range for the right-hand side of
each of these constraints.
5.11.
Reconsider the model given in Problem 5.2. While
doing what-if analysis, you learn that the estimates of the righthand sides of the two functional constraints are accurate only to
within ± 50 percent. In other words, the ranges of likely values
for these parameters are 5 to 15 for the first right-hand side and 6
to 18 for the second right-hand side.
a. After solving the original spreadsheet model, determine the shadow price for the first functional constraint by increasing its right-hand side by one and
solving again.
AS
b. Use a parameter analysis report to generate the optimal solution and total profit as the right-hand side
of the first functional constraint is incremented by 1
from 5 to 15. Use this table to estimate the allowable
range for this right-hand side, that is, the range over
which the shadow price obtained in part a is valid.
c. Repeat part a for the second functional constraint.
AS
d. Repeat part b for the second functional constraint
where its right-hand side is incremented by 1 from 6
to 18.
e. Use Solver’s sensitivity report to determine the
shadow price for each functional constraint and the
allowable range for the right-hand side of each of
these constraints.
5.12.
Consider a resource-allocation problem having the following data:
AS
Resource Usage per
Unit of Each Activity
Resource
1
2
Amount of
Resource
Available
1
1
3
8
2
1
1
4
Unit profit
$1
$2
The objective is to determine the number of units of each
activity to undertake so as to maximize the total profit.
a. Use the graphical method to solve this model.
b. Use graphical analysis to determine the shadow
price for each of these resources by solving again
Chapter 5 Problems 195
after increasing the amount of the resource available by one.
c. Use the spreadsheet model and Solver instead to do
parts a and b.
AS
d. For each resource in turn, use a parameter analysis
report to systematically generate the optimal solution and the total profit when the only change is
that the amount of that resource available increases
in increments of 1 from 4 less than the original
value up to 6 more than the original value. Use
these results to estimate the allowable range for the
amount available for each resource.
e. Use Solver’s sensitivity report to obtain the shadow
prices. Also use this report to find the range for the
amount of each resource available over which the
corresponding shadow price remains valid.
f. Describe why these shadow prices are useful
when management has the flexibility to change the
amounts of the resources being made available.
5.13.
Follow the instructions of Problem 5.12 for a resourceallocation problem that again has the objective of maximizing
total profit and that has the following data:
Resource Usage per
Unit of Each Activity
Amount of
Resource
Available
Resource
1
2
1
1
0
4
2
1
3
15
10
3
2
1
Unit profit
$3
$2
5.14.* Consider the Super Grain Corp. case study as presented
in Section 3.1, including the spreadsheet in Figure 3.1 showing
its formulation and optimal solution. Use Solver to generate the
sensitivity report. Then use this report to independently address
each of the following questions.
a. How much could the total expected number of
exposures be increased for each additional $1,000
added to the advertising budget?
b. Your answer in part a would remain valid for how
large of an increase in the advertising budget?
c. How much could the total expected number of
exposures be increased for each additional $1,000
added to the planning budget?
d. Your answer in part c would remain valid for how
large of an increase in the planning budget?
e. Would your answers in parts a and c definitely
remain valid if both the advertising budget and
planning budget were increased by $100,000 each?
f. If only $100,000 can be added to either the advertising budget or the planning budget, where should it
be added to do the most good?
g. If $100,000 must be removed from either the advertising budget or the planning budget, from which
budget should it be removed to do the least harm?
5.15.
Follow the instructions of Problem 5.14 for the continuation of the Super Grain Corp. case study as presented in Section
3.4 including the spreadsheet in Figure 3.7 showing its formulation and optimal solution.
5.16.
Consider the Union Airways problem presented in Section 3.3, including the spreadsheet in Figure 3.5 showing its formulation and optimal solution.
Management now is considering increasing the level of service provided to customers by increasing one or more of the
numbers in the rightmost column of Table 3.5 for the minimum
number of agents needed in the various time periods. To guide
them in making this decision, they would like to know what
impact this change would have on total cost.
Use Solver to generate the sensitivity report in preparation
for addressing the following questions.
a. Which of the numbers in the rightmost column of
Table 3.5 can be increased without increasing total
cost? In each case, indicate how much it can be
increased (if it is the only one being changed) without increasing total cost.
b. For each of the other numbers, how much would the
total cost increase per increase of 1 in the number?
For each answer, indicate how much the number can
be increased (if it is the only one being changed)
before the answer is no longer valid.
c. Do your answers in part b definitely remain valid if
all the numbers considered in part b are simultaneously increased by 1?
d. Do your answers in part b definitely remain valid if
all 10 numbers are simultaneously increased by 1?
e. How far can all 10 numbers be simultaneously
increased by the same amount before your answers
in part b may no longer be valid?
5.17.
Reconsider the example illustrating the use of robust
optimization that was presented in Section 5.7. Wyndor management now feels that the analysis described in this example was
overly conservative for three reasons: (1) it is unlikely that the
true values of the parameters, HD3 and HW3, will be as far as half
an hour off of the original estimate, (2) it is even more unlikely
that both estimates will turn out to simultaneously lean toward
the undesirable end of their ranges of uncertainty, and (3) there
is a bit of latitude in the constraint to compensate for violating it
by a tiny bit. Therefore, Wyndor management has asked its staff
(you) to solve the model again while using ranges of uncertainty
that are half as wide as those used in Section 5.7.
a. Apply the procedure for robust optimization with
independent parameters. What is the resulting optimal solution and how much would this increase the
total profit per week?
b. If Wyndor would need to pay a penalty of $150 per
week to the distributor if the production rates fall
below these new guaranteed minimum amounts,
should Wyndor use these new guarantees?
5.18.
Reconsider the example illustrating the use of robust
optimization that was presented in Section 5.7. Wyndor management now feels that there is uncertainty in all of the parameters
of the problem—the unit profit per door and window (PD and
PW), the hours of production time used for each door or window produced across the three plants (HD1, HW2, HD3, and HW3)
196 Chapter Five What-If Analysis for Linear Programming
and the three right-hand-sides representing the hours available
at each plant (RHS1, RHS2, and RHS3). The original estimates
along with the ranges of uncertainty are shown in the table
below. Apply the procedure for robust optimization with independent parameters to find the solution that maximizes profit
when the solution also is guaranteed to be feasible.
Parameter
Original Estimate
Range of Uncertainty
PD
$300
$250–$350
PW
$500
$400–$600
HD1
1
0.9–1.1
HW2
2
1.6–2.4
HD3
3
2.5–3.5
HW3
2
1.8–2.2
RHS1
4
3.5–4.5
RHS2
12
11–13
RHS3
18
16–20
5.19.* Reconsider Problem 5.12. Now suppose that all of the
parameters are uncertain, with ranges of uncertainty as given
in the table below. Use the procedure for robust optimization
with independent parameters to find the solution that maximizes
profit when the solution also is guaranteed to be feasible.
Resource Usage per
Unit of Each Activity
Resource
1
Amount of
Resource
Available
2
1
0.9–1.1
2.7–3.3
7.2–8.8
2
0.8–1.2
0.7–1.3
3.5–4.5
Unit profit
$0.90–$1.10
$1.75–$2.25
5.20.
Reconsider Problem 5.13. Now suppose that all of the
parameters are uncertain, with ranges of uncertainty as given
in the table below. Use the procedure for robust optimization
with independent parameters to find the solution that maximizes
profit when the solution also is guaranteed to be feasible.
Resource Usage per
Unit of Each Activity
Amount of
Resource
Available
Resource
1
1
0.8–1.2
2
0
3.6–4.4
2
0.9–1.1
2.5–3.5
13.5–16.5
3
1.6–2.4
0.7–1.3
9.5–10.5
Unit profit
$2.90–$3.10
$1.80–$2.20
AS 5.21 Reconsider the example illustrating the use of chance
constraints that was presented in Section 5.8. Suppose that after
a more careful consideration of the hours required per door and
window in plant 3, HD3 and HW3, Wyndor management has considerably narrowed the range of what these times might turn out
to be.
a. HD3 and HW3 are now considered equally likely to
be anywhere between 15 minutes below and 15
minutes above the original estimates of 3 hours
and 2 hours, respectively. Use a chance constraint
and Analytic Solver to find the solution that maximizes total profit while requiring at least a 95%
chance of satisfying the original constraint for the
third plant.
b. Repeat part a, but now require at least a 99% chance
of satisfying the constraint at the third plant.
c. Now assume HD3 and HW3 follow the normal distribution with means equal to the original estimates
of 3 hours and 2 hours, respectively, and each with
a standard deviation of 0.25 hours. Use a chance
constraint and Analytic Solver to find the solution
that maximizes total profit while requiring at least a
95% chance of satisfying the original constraint for
the third plant.
d. Repeat part c, but now require at least a 99% chance
of satisfying the original constraint at the third
plant.
AS 5.22.* Reconsider Problem 5.12, but now assume that the
amount of resource 1 available and the amount of resource 2
available are both uncertain. In particular, each follows the normal distribution, with means equal to the original estimates of
8 and 4, respectively, but each with a standard deviation of 0.5.
Use a chance constraint and Analytic Solver to find the maximum profit while requiring for each original constraint that there
be at least a 95% chance of satisfying that constraint.
AS 5.23. Reconsider Problem 5.13, but now assume that the
available amount of resource 1 and available amount of resource
2 are both uncertain.
a. If each unit of activity 1 is equally likely to use
anywhere between 1.5 and 2.5 units of resource 3,
while each unit of activity 2 is equally likely to use
anywhere between 0.7 and 1.3 units of resource 3,
then use a chance constraint and Analytic Solver
to find the maximum profit while requiring at
least a 95% chance of satisfying the original third
constraint.
b. If the amount of resource 3 used per unit of each
activity follows the normal distribution, with means
equal to the original estimates, but each with a standard deviation of 0.3, then use a chance constraint
and Analytic Solver to find the maximum profit
while requiring at least a 95% chance of satisfying
the original third constraint.
Case 5-1 Selling Soap 197
Case 5-1
Selling Soap
Reconsider the Profit & Gambit Co. advertising-mix problem
presented in Section 2.7. Recall that a major advertising campaign is being planned that will focus on three key products: a
stain remover, a liquid detergent, and a powder detergent. Management has made the following policy decisions about what
needs to be achieved by this campaign.
• Sales of the stain remover should increase by at least 3
percent.
• Sales of the liquid detergent should increase by at least 18
percent.
• Sales of the powder detergent should increase by at least 4
percent.
The spreadsheet in Figure 2.21 shows the linear programming model that was formulated for this problem. The minimum
required increases in the sales of the three products are given
in the data cells Minimum Increase (G8:G10). The changing
cells Advertising Units (C14:D14) indicate that an optimal solution for the model is to undertake four units of advertising on
television and three units of advertising in the print media. The
objective cell TotalCost (G14) shows that the total cost for this
advertising campaign would be $10 million.
After receiving this information, Profit & Gambit management now wants to analyze the trade-off between the total advertising cost and the resulting benefits achieved by increasing the
sales of the three products. Therefore, a management science
team (you) has been given the assignment of developing the
information that management will need to analyze this trade-off
and decide whether it should change any of its policy decisions
regarding the required minimum increases in the sales of the
three products. In particular, management needs some detailed
information about how the total advertising cost would change if
it were to change any or all of these policy decisions.
a. For each of the three products in turn, use graphical analysis to determine how much the total advertising cost would
change if the required minimum increase in the sales of that
product were to be increased by 1 percent (without changing
the required minimum increases for the other two products).
b. Use the spreadsheet shown in Figure 2.21 (available in
www.mhhe.com/hillier6e) to obtain the information requested
in part a.
c. For each of the three products in turn, use a parameter analysis
report to determine how the optimal solution for the model and the
resulting total advertising cost would change if the required minimum increase in the sales of that product were to be systematically
varied over a range of values (without changing the required minimum increases for the other two products). In each case, start the
range of values at 0 percent and increase by 1 percent increments
up to double the original minimum required increase.
d. Use Solver to generate the sensitivity report and indicate how the
report is able to provide the information requested in part a. Also
use the report to obtain the allowable range for the required minimum increase in the sales of each product. Interpret how each of
these allowable ranges relates to the results obtained in part c.
e. Suppose that all the original numbers in Minimum Increase
(G8:G10) were to be increased simultaneously by the same
amount. How large can this amount be before the shadow prices
provided by the sensitivity report may no longer be valid?
f. Below is the beginning of a memorandum from the management science team to Profit & Gambit management that is
intended to provide management with the information it needs
to perform its trade-off analysis. Write the rest of this memorandum based on a summary of the results obtained in the preceding parts. Present your information in clear, simple terms
that use the language of management. Avoid technical terms
such as shadow prices, allowable ranges, and so forth.
MEMORANDUM
To:
Profit & Gambit management
From:
The Management Science Team
Subject: The trade-off between advertising expenditures and increased sales
As instructed, we have been continuing our analysis of the plans for the major new advertising campaign that will focus on our spray prewash stain remover, our liquid formulation laundry
detergent, and our powder laundry detergent.
Our recent report presented our preliminary conclusions on how much advertising to do in the
different media to meet the sales goals at a minimum total cost:
Allocate $4 million to advertising on television.
Allocate $6 million to advertising in the print media.
Total advertising cost: $10 million.
(Continued)
198 Chapter Five What-If Analysis for Linear Programming
We estimate that the resulting increases in sales will be
Stain remover:
3 percent increase in sales
Liquid detergent:
18 percent increase in sales
Powder detergent:
8 percent increase in sales
You had specified that these increases should be at least 3 percent, 18 percent, and 4 percent,
respectively, so we have met the minimum levels for the first two products and substantially
exceeded it for the third.
However, you also indicated that your decisions on these minimum required increases in sales
(3 percent, 18 percent, and 4 percent) had been tentative ones. Now that we have more specific
information on what the advertising costs and the resulting increases in sales will be, you plan to
reevaluate these decisions to see if small changes might improve the trade-off between advertising cost and increased sales.
To assist you in reevaluating your decisions, we now have analyzed this trade-off for each of
the three products. Our best estimates are the following.
Case 5-2
Controlling Air Pollution
The Nori & Leets Co. is one of the major producers of steel
in its part of the world. It is located in the city of Steeltown
and is the only large employer there. Steeltown has grown and
prospered along with the company, which now employs nearly
50,000 residents. Therefore, the attitude of the townspeople
always has been, “What’s good for Nori & Leets is good for the
town.” However, this attitude is now changing; uncontrolled air
pollution from the company’s furnaces is ruining the appearance
of the city and endangering the health of its residents.
A recent stockholders’ revolt resulted in the election of a new
enlightened board of directors for the company. These directors
are determined to follow socially responsible policies, and they
have been discussing with Steeltown city officials and citizens’
groups what to do about the air pollution problem. Together they
have worked out stringent air quality standards for the Steeltown
airshed.
The three main types of pollutants in this airshed are particulate matter, sulfur oxides, and hydrocarbons. The new standards
require that the company reduce its annual emission of these pollutants by the amounts shown in the following table.
The board of directors has instructed management to have the
engineering staff determine how to achieve these reductions in
the most economical way.
The steelworks have two primary sources of pollution,
namely, the blast furnaces for making pig iron and the openhearth furnaces for changing iron into steel. In both cases, the
engineers have decided that the most effective abatement methods are (1) increasing the height of the smokestacks,1 (2) using
filter devices (including gas traps) in the smokestacks, and
(3) including cleaner, high-grade materials among the fuels
for the furnaces. Each of these methods has a technological
limit on how heavily it can be used (e.g., a maximum feasible
increase in the height of the smokestacks), but there also is
considerable flexibility for using the method at a fraction of its
technological limit.
The table below shows how much emissions (in millions of
pounds per year) can be eliminated from each type of furnace
by fully using any abatement method to its technological limit.
1
Pollutant
Particulates
Sulfur oxides
Hydrocarbons
Required Reduction in Annual
Emission Rate (million pounds)
60
150
125
Subsequent to this study, this particular abatement method
has become a controversial one. Because its effect is to reduce
ground-level pollution by spreading emissions over a greater
distance, environmental groups contend that this creates more
acid rain by keeping sulfur oxides in the air longer. Consequently,
the U.S. Environmental Protection Agency adopted new rules to
remove incentives for using tall smokestacks.
Case 5-2 Controlling Air Pollution 199
Reduction in Emission Rate from the Maximum Feasible Use of an Abatement Method
Taller Smokestacks
Pollutant
Particulates
Sulfur oxides
Hydrocarbons
Filters
Blast
Furnaces
Open-Hearth
Furnaces
Blast
Furnaces
Open-Hearth
Furnaces
Blast
Furnaces
Open-Hearth
Furnaces
12
35
37
9
42
53
25
18
28
20
31
24
17
56
29
13
49
20
For purposes of analysis, it is assumed that each method also
can be less fully used to achieve any fraction of the abatement
capacities shown in this table. Furthermore, the fractions can be
different for blast furnaces and open-hearth furnaces. For either
type of furnace, the emission reduction achieved by each method
is not substantially affected by whether or not the other methods
also are used.
After these data were developed, it became clear that no single method by itself could achieve all the required reductions. On
the other hand, combining all three methods at full capacity on
both types of furnaces (which would be prohibitively expensive
if the company’s products are to remain competitively priced) is
much more than adequate. Therefore, the engineers concluded
that they would have to use some combination of the methods,
perhaps with fractional capacities, based on their relative costs.
Furthermore, because of the differences between the blast furnaces and the open-hearth furnaces, the best combinations may
well be different for the two types of furnaces.
An analysis was conducted to estimate the total annual cost
that would be incurred by each abatement method. A method’s
annual cost includes increased operating and maintenance
expenses, as well as reduced revenue due to any loss in the efficiency of the production process caused by using the method.
The other major cost is the start-up cost (the initial capital outlay) required to install the method. To make this one-time cost
commensurable with the ongoing annual costs, the time value of
money was used to calculate the annual expenditure that would
be equivalent in value to this start-up cost.
This analysis led to the total annual cost estimates given in
the following table for using the methods at their full abatement
capacities.
Total Annual Cost from the Maximum Feasible Use of an
Abatement Method
Abatement
Method
Taller smokestacks
Filters
Better fuels
Better Fuels
Blast Furnaces
Open-Hearth
Furnaces
$8 million
7 million
11 million
$10 million
6 million
9 million
It also was determined that the cost of a method being used at
a lower level is roughly proportional to the fraction of the abatement capacity (given in the preceding table) that is achieved.
Thus, for any given fraction achieved, the total annual cost
would be roughly that fraction of the corresponding quantity in
the cost table.
The stage now is set to develop the general framework of
the company’s plan for pollution abatement. This plan needs to
specify which types of abatement methods will be used and at
what fractions of their abatement capacities for (1) the blast furnaces and (2) the open-hearth furnaces.
You have been asked to head a management science team
to analyze this problem. Management wants you to begin by
determining which plan would minimize the total annual cost
of achieving the required reductions in annual emission rates for
the three pollutants.
a. Identify verbally the components of a linear programming
model for this problem.
b. Display the model on a spreadsheet.
c. Obtain an optimal solution and generate the sensitivity report.
Management now wants to conduct some what-if analysis
with your help. Since the company does not have much prior
experience with the pollution abatement methods under consideration, the cost estimates given in the third table are fairly
rough, and each one could easily be off by as much as 10 percent
in either direction. There also is some uncertainty about the values given in the second table, but less so than for the third table.
By contrast, the values in the first table are policy standards and
so are prescribed constants.
However, there still is considerable debate about where to set
these policy standards on the required reductions in the emission
rates of the various pollutants. The numbers in the first table
actually are preliminary values tentatively agreed upon before
learning what the total cost would be to meet these standards.
Both the city and company officials agree that the final decision on these policy standards should be based on the trade-off
between costs and benefits. With this in mind, the city has concluded that each 10 percent increase in the policy standards over
the current values (all the numbers in the first table) would be
worth $3.5 million to the city. Therefore, the city has agreed to
reduce the company’s tax payments to the city by $3.5 million
for each 10 percent increase in the policy standards (up to 50
percent) that is accepted by the company.
Finally, there has been some debate about the relative values
of the policy standards for the three pollutants. As indicated in
the first table, the required reduction for particulates now is less
than half of that for either sulfur oxides or hydrocarbons. Some
have argued for decreasing this disparity. Others contend that
200 Chapter Five What-If Analysis for Linear Programming
an even greater disparity is justified because sulfur oxides and
hydrocarbons cause considerably more damage than particulates. Agreement has been reached that this issue will be reexamined after information is obtained about which trade-offs in
policy standards (increasing one while decreasing another) are
available without increasing the total cost.
d. Identify the parameters of the linear programming model that
should be classified as sensitive parameters. Make a resulting recommendation about which parameters should be estimated more closely, if possible.
e. Analyze the effect of an inaccuracy in estimating each cost
parameter given in the third table. If the true value were 10
percent less than the estimated value, would this change the
optimal solution? Would it change if the true value were 10
percent more than the estimated value? Make a resulting recommendation about where to focus further work in estimating the cost parameters more closely.
f. For each pollutant, specify the rate at which the total cost of
an optimal solution would change with any small change in
the required reduction in the annual emission rate of the pollutant. Also specify how much this required reduction can be
changed (up or down) without affecting the rate of change in
the total cost.
g. For each unit change in the policy standard for particulates
given in the first table, determine the change in the opposite
direction for sulfur oxides that would keep the total cost of
an optimal solution unchanged. Repeat this for hydrocarbons
instead of sulfur oxides. Then do it for a simultaneous and
equal change for both sulfur oxides and hydrocarbons in the
opposite direction from particulates.
h. Letting θ denote the percentage increase in all the policy standards given in the first table, use a parameter analysis report
to systematically find an optimal solution and the total cost for
the revised linear programming problem for each θ = 10, 20,
30, 40, 50. Considering the tax incentive offered by the city,
use these results to determine which value of θ (including the
option of θ = 0) should be chosen by the company to minimize its total cost of both pollution abatement and taxes.
i. For the value of θ chosen in part h, generate the sensitivity
report and repeat parts f and g so that the decision makers
can make a final decision on the relative values of the policy
standards for the three pollutants.
Case 5-3
Farm Management
The Ploughman family owns and operates a 640-acre farm that has
been in the family for several generations. The Ploughmans always
have had to work hard to make a decent living from the farm and
have had to endure some occasional difficult years. Stories about
earlier generations overcoming hardships due to droughts, floods,
and so forth, are an important part of the family history. However,
the Ploughmans enjoy their self-reliant lifestyle and gain considerable satisfaction from continuing the family tradition of successfully living off the land during an era when many family farms are
being abandoned or taken over by large agricultural corporations.
John Ploughman is the current manager of the farm, while
his wife Eunice runs the house and manages the farm’s finances.
John’s father, Grandpa Ploughman, lives with them and still puts
in many hours working on the farm. John and Eunice’s children,
Frank, Phyllis, and Carl, also are given heavy chores before and
after school.
The entire family can produce a total of 4,000 person-hours’
worth of labor during the winter and spring months and 4,500
person-hours during the summer and fall. If any of these personhours are not needed, Frank, Phyllis, and Carl will use them to
work on a neighboring farm for $7/hour during the winter and
spring months and $7.70/hour during the summer and fall.
The farm supports two types of livestock, dairy cows and
laying hens, as well as three crops: soybeans, corn, and wheat.
(All three are cash crops, but the corn also is a feed crop for the
cows and the wheat also is used for chicken feed.) The crops
are harvested during the late summer and fall. During the winter
months, John, Eunice, and Grandpa make a decision about the
mix of livestock and crops for the coming year.
Currently, the family has just completed a particularly successful harvest that has provided an investment fund of $28,000
that can be used to purchase more livestock. (Other money is
available for ongoing expenses, including the next planting of
crops.) The family currently has 30 cows valued at $49,000 and
2,000 hens valued at $7,000. They wish to keep all this livestock
and perhaps purchase more. Each new cow would cost $2,100,
and each new hen would cost $4.20.
Over a year’s time, the value of a herd of cows will decrease
by about 10 percent and the value of a flock of hens will decrease
by about 25 percent due to aging.
Each cow will require two acres of land for grazing and 10
person-hours of work per month, while producing a net annual
cash income of $1,190 for the family. The corresponding figures
for each hen are no significant acreage, 0.05 person-hours per
month, and an annual net cash income of $5.95. The chicken
house can accommodate a maximum of 5,000 hens, and the size
of the barn limits the herd to a maximum of 42 cows.
For each acre planted in each of the three crops, the first table
on the next page gives the number of person-hours of work that
will be required during the first and second halves of the year, as
well as a rough estimate of the crop’s net value (in either income
or savings in purchasing feed for the livestock).
To provide much of the feed for the livestock, John wants to
plant at least one acre of corn for each cow in the coming year’s
Case 5-3 Farm Management 201
herd and at least 0.05 acre of wheat for each hen in the coming
year’s flock.
John, Eunice, and Grandpa now are discussing how much
acreage should be planted in each of the crops and how many
cows and hens to have for the coming year. Their objective is to
maximize the family’s monetary worth at the end of the coming year (the sum of the net income from the livestock for the
coming year plus the net value of the crops for the coming year
plus what remains from the investment fund plus the value of the
livestock at the end of the coming year plus income from working on a neighboring farm minus living expenses of $56,000 for
the year).
Soybeans Corn Wheat
Winter and spring, person-hours
1.0
0.9
0.6
Summer and fall, person-hours
1.4
1.2
0.7
$98
$84
$56
Net value
a. Identify verbally the components of a linear programming
model for this problem.
b. Display the model on a spreadsheet.
c. Obtain an optimal solution and generate the sensitivity
report. What does the model predict regarding the family’s
monetary worth at the end of the coming year?
d. Find the allowable range for the net value per acre planted for
each of the three crops.
The above estimates of the net value per acre planted in each
of the three crops assumes good weather conditions. Adverse
weather conditions would harm the crops and greatly reduce the
resulting value. The scenarios particularly feared by the family
are a drought, a flood, an early frost, both a drought and an early
frost, and both a flood and an early frost. The estimated net values for the year under these scenarios are shown next.
Net Value per Acre Planted
Scenario
Drought
Flood
Early frost
Drought and early frost
Flood and early frost
Soybeans
Corn
Wheat
−$14
−$21
0
21
28
$14
70
56
42
−21
−28
−14
14
14
7
e. Find an optimal solution under each scenario after making
the necessary adjustments to the linear programming model
formulated in part b. In each case, what is the prediction
regarding the family’s monetary worth at the end of the year?
f. For the optimal solution obtained under each of the six scenarios (including the good weather scenario considered in
parts a-d), calculate what the family’s monetary worth would
be at the end of the year if each of the other five scenarios
occurs instead. In your judgment, which solution provides
the best balance between yielding a large monetary worth
under good weather conditions and avoiding an overly small
monetary worth under adverse weather conditions?
Grandpa has researched what the weather conditions were in
past years as far back as weather records have been kept and
obtained the data shown in the table below. With these data,
the family has decided to use the following approach to making
its planting and livestock decisions. Rather than the optimistic
approach of assuming that good weather conditions will prevail
(as done in parts a-d), the average net value under all weather
conditions will be used for each crop (weighting the net values
under the various scenarios by the frequencies in the following
table).
Scenario
Frequency
Good weather
40%
Drought
20
Flood
10
Early frost
15
Drought and early frost
10
Flood and early frost
5
g. Modify the linear programming model formulated in part b
to fit this new approach.
h. Repeat part c for this modified model.
i. Use a shadow price obtained in part h to analyze whether
it would be worthwhile for the family to obtain a bank loan
with a 10 percent interest rate to purchase more livestock
now beyond what can be obtained with the $28,000 from the
investment fund.
j. For each of the three crops, use the sensitivity report obtained
in part h to identify how much latitude for error is available in
estimating the net value per acre planted for that crop without
changing the optimal solution. Which two net values need
to be estimated most carefully? If both estimates are incorrect simultaneously, how close do the estimates need to be
to guarantee that the optimal solution will not change? Use a
two-way parameter analysis report to systematically generate
the optimal monetary worth as these two net values are varied simultaneously over ranges that go up to twice as far from
the estimates as needed to guarantee that the optimal solution
will not change.
This problem illustrates a kind of situation that is frequently
faced by various kinds of organizations. To describe the situation in general terms, an organization faces an uncertain future
where any one of a number of scenarios may unfold. Which one
will occur depends on conditions that are outside the control of
the organization. The organization needs to choose the levels of
various activities, but the unit contribution of each activity to
the overall measure of performance is greatly affected by which
scenario unfolds. Under these circumstances, what is the best
mix of activities?
k. Think about specific situations outside of farm management
that fit this description. Describe one.
202 Chapter Five What-If Analysis for Linear Programming
Case 5-4
Assigning Students to Schools (Revisited)
Reconsider Case 3-5. The Springfield School Board still has
the policy of providing busing for all middle school students
who must travel more than approximately a mile. Another current policy is to allow splitting residential areas among multiple
schools if this will reduce the total busing cost. (This latter policy will be reversed in Case 7-3.) However, before adopting a
busing plan based on part a of Case 3-5, the school board now
wants to conduct some what-if analysis.
a. If you have not already done so for part a of Case 3-5, formulate and solve a linear programming model for this problem
on a spreadsheet.
b. Use Solver to generate the sensitivity report.
One concern of the school board is the ongoing road construction in area 6. These construction projects have been delaying traffic considerably and are likely to affect the cost of busing
students from area 6, perhaps increasing costs as much as 10
percent.
c. Use the sensitivity report to check how much the busing cost
from area 6 to school 1 can increase (assuming no change
in the costs for the other schools) before the current optimal solution would no longer be optimal. If the allowable
increase is less than 10 percent, use Solver to find the new
optimal solution with a 10 percent increase.
d. Repeat part c for school 2 (assuming no change in the costs
for the other schools).
e. Now assume that the busing cost from area 6 would increase
by the same percentage for all the schools. Use the sensitivity
report to determine how large this percentage can be before
the current optimal solution might no longer be optimal. If
the allowable increase is less than 10 percent, use Solver to
find the new optimal solution with a 10 percent increase.
The school board has the option of adding portable classrooms to increase the capacity of one or more of the middle
schools for a few years. However, this is a costly move that the
board would only consider if it would significantly decrease
busing costs. Each portable classroom holds 20 students and
has a leasing cost of $2,500 per year. To analyze this option,
the school board decides to assume that the road construction in
area 6 will wind down without significantly increasing the busing costs from that area.
f. For each school, use the corresponding shadow price from
the sensitivity report to determine whether it would be worthwhile to add any portable classrooms.
g. For each school where it is worthwhile to add any portable
classrooms, use the sensitivity report to determine how many
could be added before the shadow price would no longer be
valid (assuming this is the only school receiving portable
classrooms).
h. If it would be worthwhile to add portable classrooms to more
than one school, use the sensitivity report to determine the
combinations of the number to add for which the shadow
prices definitely would still be valid. Then use the shadow
prices to determine which of these combinations is best in
terms of minimizing the total cost of busing students and
leasing portable classrooms. Use Solver for finding the corresponding optimal solution for assigning students to schools.
i. If part h was applicable, modify the best combination of portable classrooms found there by adding one more to the school
with the most favorable shadow price. Use Solver to find
the corresponding optimal solution for assigning students to
schools and to generate the corresponding sensitivity report.
Use this information to assess whether the plan developed in
part h is the best one available for minimizing the total cost of
busing students and leasing portables. If not, find the best plan.
Additional Case
Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website,
cases.ivey.uwo.ca/cases, in the segment of the CaseMate area
designated for this book.
Chapter SIX
Network Optimization
Problems
Learning Objectives
After completing this chapter, you should be able to
1. Formulate network models for various types of network optimization problems.
2. Describe the characteristics of minimum-cost flow problems, maximum flow problems, and
shortest path problems.
3. Identify some areas of application for these types of problems.
4. Identify several categories of network optimization problems that are special types of
minimum-cost flow problems.
5. Formulate and solve a spreadsheet model for a minimum-cost flow problem, a maximum
flow problem, or a shortest path problem from a description of the problem.
This chapter focuses on how to analyze networks, so let's begin with a description of what
they are. A network can be defined as a system with a number of components where there are
direct connections between various pairs of components. These connections might be actual
physical connections, such as the wires in an electrical network. In other cases, the network
might be a diagram that depicts the locations of the components of the system and then uses
lines between certain pairs of components to identify direct relationships between these components. For example, if you refer back to Section 3.5, Figure 3.9 shows a network that depicts
a distribution system with two types of components (factories and customers) where the lines
from factories to customers identify the feasible shipping routes.
Networks arise in numerous settings and in a variety of guises. Transportation, electrical,
and communication networks pervade our daily lives. Network representations also are widely
used for problems in such diverse areas as production, distribution, project planning, facilities
location, resource management, and financial planning—to name just a few examples. In fact,
a network representation provides such a powerful visual and conceptual aid for portraying
the relationships between the components of systems that it is used in virtually every field of
scientific, social, and economic endeavor.
This chapter focuses on how to use spreadsheet models to optimize the operation of certain
kinds of networks. All of these networks are designed to provide flow of some type from certain
points of origin along various permissible routes to certain destinations. For example, this flow
might involve the shipment of goods or resources through the network. Or perhaps the objective is to identify the optimal route through the network, in which case the flow simply involves
travel along that route. For each type of network, we will show how a spreadsheet model can
formulate and solve the problem of how to operate the network in an optimal manner.
One of the most exciting developments in management science in recent decades has been
the unusually rapid advance in both the methodology and application of network optimization
problems. A number of algorithmic breakthroughs have had a major impact, as have ideas
from computer science concerning data structures and efficient data manipulation. Consequently, algorithms and software now are available and are being used to solve huge problems
on a routine basis that would have been completely intractable a few decades ago. However,
we will only consider small examples in this chapter.
203
204 Chapter Six Network Optimization Problems
This chapter presents the network optimization problems that have been particularly helpful in dealing with managerial issues. We focus on the nature of these problems and their
applications rather than on the technical details and the algorithms used to solve the problems.
Using spreadsheet models and Solver will enable us to maintain this focus.
You already have seen some examples of network optimization problems in Chapter 3. In
particular, transportation problems (described in Section 3.5) have a network representation
that is shown in Figure 3.9 already cited earlier. A typical transportation problem involves
shipping some product from certain factories to certain customers, where the objective is to
identify the shipping plan that minimizes the total cost of these shipments. Assignment problems (Section 3.6) have a similar network representation (as described further in Chapter 15 at
www.mhhe.com/Hillier6e). Therefore, both transportation problems and assignment problems are simple types of network optimization problems.
Like transportation problems and assignment problems, many other network optimization
problems (including all the types considered in this chapter) also are special types of linear
programming problems. Consequently, after formulating a spreadsheet model for these problems, they can be readily solved by Solver.
Section 6.1 discusses an especially important type of network optimization problem called
a minimum-cost flow problem. A typical application involves minimizing the cost of shipping
goods through a distribution network. Thus, this problem is similar to a transportation problem
except now there are some intermediate points (e.g., warehouses) in the distribution network.
Section 6.3 presents maximum flow problems, which are concerned with such issues as
how to maximize the flow of goods through a distribution network. Section 6.2 lays the
groundwork by introducing a case study of a maximum flow problem.
Section 6.4 considers shortest path problems. In their simplest form, the objective is to
find the shortest route through a network between two locations.
A supplement to this chapter at www.mhhe.com/Hillier6e discusses minimum spanningtree problems, which are concerned with minimizing the cost of providing either direct or indirect connections between all users of a system. This is the only network optimization problem
considered in this book that is not, in fact, a special type of linear programming problem.
6.1
MINIMUM-COST FLOW PROBLEMS
Before describing the general characteristics of minimum-cost flow problems, let us first look
at an example. For clarity, this first example is a tiny one, but otherwise it is typical of the
sometimes huge minimum-cost flow problems that arise in practice.
An Example: The Distribution Unlimited Co. Problem
The Distribution Unlimited Co. has two factories producing a product that needs to be
shipped to two warehouses. Here are some details.
Factory 1 is producing 80 units.
Factory 2 is producing 70 units.
Warehouse 1 needs 60 units.
Warehouse 2 needs 90 units.
The objective is to minimize
the total shipping cost
through the distribution
network.
(Each unit corresponds to a full truckload of the product.)
Figure 6.1 shows the distribution network available for shipping this product, where F1
and F2 are the two factories, W1 and W2 are the two warehouses, and DC is a distribution
center. The arrows show feasible shipping lanes. In particular, there is a rail link from Factory 1 to Warehouse 1 and another from Factory 2 to Warehouse 2. (Any amounts can be
shipped along these rail links.) In addition, independent truckers are available to ship up to
50 units from each factory to the distribution center, and then to ship up to 50 units from the
distribution center to each warehouse. (Whatever is shipped to the distribution center must
subsequently be shipped on to the warehouses.) Management’s objective is to determine the
shipping plan (how many units to ship along each shipping lane) that will minimize the total
shipping cost.
6.1 Minimum-Cost Flow Problems 205
FIGURE 6.1
The distribution network
for the Distribution
Unlimited Co. problem,
where each feasible shipping lane is represented
by an arrow.
80 units
produced
F1
W1
60 units
needed
W2
90 units
needed
DC
70 units
produced
F2
If you have read the description of transportation problems in Section 3.5 (as illustrated in
Figure 3.9), you will notice that this minimum-cost flow problem resembles a transportation
problem. However, the key difference is that this example includes an intermediate destination (the distribution center), which does not fit the definition of a transportation problem.
Many applications include such intermediate destinations, which makes minimum-cost flow
problems a valuable generalization of transportation problems.
The shipping costs differ considerably among these shipping lanes. The cost per unit shipped
through each lane is shown above the corresponding arrow in the network in Figure 6.2.
To make the network less crowded, the problem usually is presented even more compactly, as shown in Figure 6.3. The number in square brackets next to the location of each
facility indicates the net number of units (outflow minus inflow) generated there. Thus,
the number of units terminating at each warehouse is shown as a negative number. The
number at the distribution center is 0 since the number of units leaving minus the number
of units arriving must equal 0. The number on top of each arrow shows the unit shipping
cost along that shipping lane. Any number in square brackets underneath an arrow gives
[80]
$700/unit
60 units
needed
$3
DC
[5
0
0
t
ni x.]
a
/u
00 s m
it
un
[5
[0]
$2
0
un 0/u
its nit
m
ax
.]
W1
00
$3 0]
[5
F1
0
[5
F2
FIGURE 6.2
[5
t
ni x.]
a
/u
m
00
$4 nits
u
0
[5
$5
0
un 0/u
its nit
m
ax
.]
0]
$5
00
00
$4 0]
[5
DC
70 units
produced
W1
$2
[5 00
0]
F1
80 units
produced
[–60]
$700
$1,000/unit
F2
W2
90 units
needed
The data for the distribution network for the ­Distribution
Unlimited Co. problem.
[70]
FIGURE 6.3
$1,000
W2
[–90]
A network model for the Distribution
Unlimited Co. problem as a minimum-cost
flow problem.
An Application Vignette
Hewlett-Packard (HP) offers many innovative products to meet
the diverse needs of more than 1 billion customers. The breadth
of its product offering has helped the company achieve unparalleled market reach. However, offering multiple similar products also can cause serious problems—including confusing
sales representatives and customers—that can adversely affect
the revenue and costs for any particular product. Therefore, it is
important to find the right balance between too much product
variety and too little.
With this in mind, HP top management made managing
product variety a strategic business priority. HP has been a
leader in applying management science to its important business problems for decades, so it was only natural that many
of the company’s top management scientists were called on to
address this problem as well.
The heart of the methodology that was developed to address
this problem involved formulating and applying a network optimization model. After excluding proposed products that do not
have a sufficiently high return on investment, the remaining
Figure 6.3 illustrates how
a minimum-cost flow
problem can be completely
depicted by a network.
proposed products can be envisioned as flows through a network that can help fill some of the projected orders on the righthand side of the network. The resulting model is a special type
of minimum cost flow problem (related to the special type discussed in the next two sections).
Following its implementation by the beginning of 2005, this
application of a minimum cost flow problem had a dramatic
impact in enabling HP businesses to increase operational focus
on their most critical products. This yielded companywide profit
improvements of over $500 million between 2005 and 2008,
and then about $180 million annually thereafter. It also yielded
a variety of important qualitative benefits for HP.
These dramatic results led to HP winning the prestigious
first prize in the 2009 Franz Edelman Award for Achievement in
Operations Research and the Management Sciences.
Source: J. Ward and 20 co-authors, “HP Transforms Product Portfolio
Management with Operations Research,” Interfaces 40, no. 1
(January–February 2010), pp. 17–32. (A link to this article is provided
at www.mhhe.com/Hillier6e.)
the maximum number of units that can be shipped along that shipping lane. (The absence
of a number in square brackets underneath an arrow implies that there is no limit on the
shipping amount there.) This network provides a complete representation of the problem,
including all the necessary data, so it constitutes a network model for this minimum-cost
flow problem.
Since this is such a tiny problem, you probably can see what the optimal solution must be.
(Try it.) This solution is shown in Figure 6.4, where the shipping amount along each shipping
lane is given in parentheses. (To avoid confusion, we delete the unit shipping costs and shipping capacities in this figure.) Combining these shipping amounts with the unit shipping costs
given in Figures 6.2 and 6.3, the total shipping cost for this solution (when starting by listing
the costs from F1, then from F2, and then from DC) is
Total shipping cost = 30($700) + 50($300) + 30($500) + 40($1,000)
+ 30($200) + 50($400)
= $117,000
[80]
[–60]
(30)
F1
W1
0)
0)
(5
The optimal solution for
the Distribution Unlimited Co. problem, where
the shipping amounts are
shown in parentheses over
the arrows.
(3
FIGURE 6.4
[0]
DC
(3
0)
0)
(5
F2
[70]
206
(40)
W2
[–90]
6.1 Minimum-Cost Flow Problems 207
General Characteristics
This example possesses all the general characteristics of any minimum-cost flow problem.
Before summarizing these characteristics, here is the terminology you will need.
Terminology
A supply node has net
flow going out whereas a
demand node has net flow
coming in.
1. The model for any minimum-cost flow problem is represented by a network with flow
passing through it.
2. The circles in the network are called nodes.
3. Each node where the net amount of flow generated (outflow minus inflow) is a fixed
­positive number is a supply node. (Thus, F1 and F2 are the supply nodes in Figure 6.3.)
4. Each node where the net amount of flow generated is a fixed negative number is a demand
node. (Consequently, W1 and W2 are the demand nodes in the example.)
5. Any node where the net amount of flow generated is fixed at zero is a transshipment node.
(Thus, DC is the transshipment node in the example.) Having the amount of flow out of the
node equal the amount of flow into the node is referred to as conservation of flow.
6. The arrows in the network are called arcs.
7. The maximum amount of flow allowed through an arc is referred to as the capacity of
that arc.
Using this terminology, the general characteristics of minimum-cost flow problems (the
model for this type of problem) can be described in terms of the following assumptions.
Assumptions of a Minimum-Cost Flow Problem
Since the arrowhead on an
arc indicates the direction
in which flow is allowed,
a pair of arcs pointing in
opposite directions is used
if flow can occur in both
directions.
The objective is to minimize the total cost of supplying the demand nodes.
1. At least one of the nodes is a supply node, so each one specifies its own fixed positive
­number for the net amount of flow generated there.
2. At least one of the other nodes is a demand node, so each one specifies its own fixed
­negative number for the net amount of flow generated there.
3. All the remaining nodes are transshipment nodes, so each one specifies a fixed value of
zero for the net amount of flow generated there.
4. Flow through an arc is only allowed in the direction indicated by the arrowhead, where
the maximum amount of flow is given by the capacity of that arc. (If flow can occur
in both directions, this would be represented by a pair of arcs pointing in opposite
directions.)
5. The network has enough arcs with sufficient capacity to enable all the flow generated at
the supply nodes to reach all the demand nodes.
6. The cost of the flow through each arc is proportional to the amount of that flow, where the
cost per unit flow is known.
7. The objective is to minimize the total cost of sending the available supply through the network to satisfy the given demand. (An alternative objective is to maximize the total profit
from doing this.)
A solution for this kind of problem needs to specify how much flow is going through
each arc. To be a feasible solution, the amount of flow through each arc cannot exceed the
capacity of that arc and the net amount of flow generated at each node must equal the specified amount for that node. The following property indicates when the problem will have
feasible solutions.
Feasible Solutions Property: Under the above assumptions, a minimum-cost flow problem will
have feasible solutions if and only if the sum of the supplies from its supply nodes equals the
sum of the demands at its demand nodes.
Note that this property holds for the Distribution Unlimited Co. problem, because the sum of
its supplies is 80 + 70 = 150 and the sum of its demands is 60 + 90 = 150.
Applications occasionally do arise where the sum of the supplies from the supply nodes
does not equal the sum of the demands at its demand nodes (an “unbalanced problem”). For
208 Chapter Six Network Optimization Problems
example, suppose that the sum of the supplies exceeds the sum of the demands by some
amount (the “extra supply”). In this case, the “supplies” actually represent the maximum
amounts that could be supplied rather than fixed amounts, so part of the problem is to determine exactly how much should be produced at each supply node to exactly meet the demands
at the demand nodes. Such a problem no longer exactly fits the definition of a minimumcost flow problem because it violates assumption 1. Nevertheless, Solver can still solve the
problem by using the corresponding spreadsheet model that includes ≤ constraints for the
amounts produced at each supply node. Alternatively, the problem can be converted back
into a minimum-cost flow problem by adding a “dummy demand node” to the network where
the demand at this node equals the “extra supply” at the supply nodes. By adding an arc with
a zero cost per unit flow (and a large arc capacity) from each supply node to the dummy
demand node, the network now becomes a minimum-cost flow problem because it now satisfies the feasible solutions property.
For many applications of minimum-cost flow problems, management desires a solution with integer values for all the flow quantities (e.g., integer numbers of full truckloads
along each shipping lane). The model does not include any constraints that require this for
feasible solutions. Fortunately, such constraints are not needed because of the following
property.
Integer Solutions Property: As long as all its supplies, demands, and arc capacities have integer values, any minimum-cost flow problem with feasible solutions is guaranteed to have an
optimal solution with integer values for all its flow quantities.
See in Figure 6.3 that all the assumptions needed for this property to hold are satisfied for
the Distribution Unlimited Co. problem. In particular, all the supplies (80 and 70), demands
(60 and 90), and arc capacities (50) have integer values. Therefore, all the flow quantities in
the optimal solution given in Figure 6.4 (30 three times, 50 two times, and 40) have integer
values. This ensures that only full truckloads will be shipped into and out of the distribution
center. (Remember that each unit corresponds to a full truckload of the product.)
Now let us see how to obtain an optimal solution for the Distribution Unlimited Co. problem by formulating a spreadsheet model and then applying Solver.
Using Excel to Formulate and Solve Minimum-Cost
Flow Problems
Figure 6.5 shows a spreadsheet model that is based directly on the network representation of
the problem in Figure 6.3. The arcs are listed in columns B and C, along with their capacities (unless unlimited) in column F and their costs per unit flow in column G. The changing
cells Ship (D4:D9) show the flow amounts through these arcs and the objective cell TotalCost
(D11) provides the total cost of this flow by using the equation
​D11 = SUMPRODUCT(Ship, UnitCost)​
Capacity constraints like
these are needed in any
minimum-cost flow problem that has any arcs with
limited capacity.
Any minimum-cost flow
problem needs net flow
constraints like this for
every node.
Excel Tip: SUMIF(A, B,
C) adds up each entry in the
range C for which the corresponding entry in range
A equals B. This function is
especially useful in network
problems for calculating the
net flow generated at a node.
The first set of constraints in the Solver Parameters box, D5:D8 ≤ Capacity (F5:F8), ensures
that the arc capacities are not exceeded.
Similarly, Column I lists the nodes, column J calculates the actual net flow generated at
each node (given the flows in the changing cells), and column L specifies the net amount of
flow that needs to be generated at each node. Thus, the second set of constraints in the Solver
Parameters box is NetFlow (J4:J8) = SupplyDemand (L4:L8), requiring that the actual net
amount of flow generated at each node must equal the specified amount.
The equations entered into NetFlow (J4:J8) use the difference of two SUMIF functions to
calculate the net flow (outflow minus inflow) generated at each node. In particular, the first
SUMIF function calculates the flow leaving the node and the second one calculates the flow
entering the node. (If you haven't used SUMIF functions previously, the nearby Excel tip
defines what this kind of function does.)
For example, consider the F1 node (I4) and the calculation in J4 of the net flow for this
node. The equation for this net flow is
​​J4 = SUMIF​(From, I4, Ship)​ − SUMIF​(​​To, I4, Ship​)​​​​
6.1 Minimum-Cost Flow Problems 209
FIGURE 6.5
A spreadsheet model for the Distribution Unlimited Co. minimum-cost flow problem, including the objective cell TotalCost (D11)
and the other output cells NetFlow (J4:J8), as well as the equations entered into these cells and the other specifications needed to
set up the model. The changing cells Ship (D4:D9) show the optimal shipping quantities through the distribution network obtained
by Solver.
A
1
B
C
D
E
F
G
H
K
L
I
J
Unit Cost
Nodes
Net Flow
$700
F1
80
=
80
70
Distribution Unlimited Co. Minimum Cost Flow Problem
2
3
From
To
Ship
4
F1
W1
30
Capacity
Supply/Demand
5
F1
DC
50
≤
50
$300
F2
70
=
6
DC
W1
30
≤
50
$200
DC
0
=
0
7
DC
W2
50
≤
50
$400
W1
-60
=
-60
≤
50
$500
W2
-90
=
-90
8
F2
DC
30
9
F2
W2
40
$1,000
10
11
Total Cost
$117,000
Solver Parameters
Set Objective Cell: TotalCost
To: Min
By Changing Variable Cells:
Ship
Subject to the Constraints:
D5:D8 <= Capacity
NetFlow = SupplyDemand
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
J
Range Name
Cells
Capacity
From
NetFlow
Nodes
Ship
SupplyDemand
To
TotalCost
UnitCost
F5:F8
B4:B9
J4:J8
I4:I8
D4:D9
L4:L8
C4:C9
D11
G4:G9
11
3
Net Flow
4
=SUMIF(From,I4,Ship)-SUMIF(To,I4,Ship)
5
=SUMIF(From,I5,Ship)-SUMIF(To,I5,Ship)
6
=SUMIF(From,I6,Ship)-SUMIF(To,I6,Ship)
7
=SUMIF(From,I7,Ship)-SUMIF(To,I7,Ship)
8
=SUMIF(From,I8,Ship)-SUMIF(To,I8,Ship)
C
D
Total Cost
=SUMPRODUCT(Ship,UnitCost)
SUMIF(From,I4,Ship) sums each individual entry in Ship (D4:D9) if that entry is in a row
where the entry in From (B4:B9) is the same as in I4. Since I4 = F1 and the only rows that
have F1 in the From column are rows 4 and 5, the sum in the Ship column is only over these
same rows, so this sum is D4 + D5. Similarly, SUMIF(To,I4,Ship) sums each individual entry
in Ship (D4:D9) if that entry is in a row where the entry in To (C4:C9) is the same as in I4.
However, F1 never appears in the To column, so this sum is 0. Therefore, the overall equation
for J4 yields
​J4 = D4 + D5 − 0 = 30 + 50 − 0 = 80,​
which is the net flow generated at the F1 node.
While it appears more complicated to use the SUMIF function rather than just entering
J4 = D4 + D5, J5 = D8 + D9, J6 = D6 + D7 − D5 − D8, and so on, it is actually simpler. The
SUMIF formula only needs to be entered once (in cell J4). It can then be copied down into the
remaining cells in NetFlow (J5:J8). For a problem with many nodes, this is much quicker and
(perhaps more significantly) less prone to error. In a large problem, it is all too easy to miss
an arc when determining which cells in the Ship column to add and subtract to calculate the
net flow for a given node.
The first Solver option specifies that the flow amounts cannot be negative. The second
acknowledges that this is still a linear programming problem.
Running Solver gives the optimal solution shown in Ship (D4:D9) in Figure 6.5. This is the
same solution as displayed in Figure 6.4.
210 Chapter Six Network Optimization Problems
Solving Large Minimum-Cost Flow Problems More Efficiently
network simplex method
The network simplex
method can solve much
larger minimum-cost flow
problems (sometimes with
millions of nodes and
arcs) than can the simplex
method used by Solver.
Because minimum-cost flow problems are a special type of linear programming problem, and
the simplex method can solve any linear programming problem, it also can solve any minimumcost flow problem in the standard way. For example, Solver uses the simplex method to solve
this type (or any other type) of linear programming problem. This works fine for small problems, like the Distribution Unlimited Co. problem, and for considerably larger ones as well.
Therefore, the approach illustrated in Figure 6.5 will serve you well for any minimum-cost
flow problem encountered in this book and for many that you will encounter subsequently.
However, we should mention that a different approach is sometimes needed in practice to
solve really big problems. Because of the special form of minimum-cost flow problems, it is
possible to greatly streamline the simplex method to solve them far more quickly. In particular, rather than going through all the algebra of the simplex method, it is possible to execute
the same steps far more quickly by working directly with the network for the problem.
This streamlined version of the simplex method is called the network simplex method.
The network simplex method can solve some huge problems that are much too large for the
simplex method.
Like the simplex method, the network simplex method not only finds an optimal solution but also can be a valuable aid to managers in conducting the kinds of what-if analyses
described in Chapter 5.
Many companies now use the network simplex method to solve their minimum-cost flow
problems. Some of these problems are huge, with many tens of thousands of nodes and arcs.
Occasionally, the number of arcs will even be far larger, perhaps into the millions.
Although Solver does not, other commercial software packages for linear programming
commonly include the network simplex method.
Another important advance in fairly recent years has been the development of excellent
graphical interfaces for modeling minimum-cost flow problems. These interfaces make the
design of the model and the interpretation of the output of the network simplex method completely visual and intuitive with no mathematics involved. This is very helpful for managerial
decision making.
Some Applications
Probably the most important kind of application of minimum-cost flow problems is to the
operation of a distribution network, such as the one depicted in Figures 6.1–6.4 for the Distribution Unlimited Co. problem. As summarized in the first row of Table 6.1, this kind of
application involves determining a plan for shipping goods from their sources (factories, etc.)
to intermediate storage facilities (as needed) and then on to the customers.
For some applications of minimum-cost flow problems, all the transshipment nodes are
processing facilities rather than intermediate storage facilities. This is the case for solid waste
management, as indicated in Table 6.1. Here, the flow of materials through the network begins
at the sources of the solid waste, then goes to the facilities for processing these waste materials into a form suitable for landfill, and then sends them on to the various landfill locations.
However, the objective still is to determine the flow plan that minimizes the total cost, where
the cost now is for both shipping and processing.
TABLE 6.1
Typical Kinds of Applications of Minimum-Cost Flow Problems
Kind of Application
Supply Nodes
Transshipment Nodes
Demand Nodes
Operation of a distribution network
Solid waste management
Operation of a supply network
Sources of goods
Sources of solid waste
Vendors
Intermediate storage facilities
Processing facilities
Intermediate warehouses
Coordinating product mixes at plants
Plants
Sources of cash at a
specific time
Production of a specific product
Customers
Landfill locations
Processing facilities
Market for a specific
product
Needs for cash at a
specific time
Cash flow management
Short-term investment options
6.1 Minimum-Cost Flow Problems 211
In other applications, the demand nodes might be processing facilities. For example, in the
third row of Table 6.1, the objective is to find the minimum-cost plan for obtaining supplies
from various possible vendors, storing these goods in warehouses (as needed), and then shipping the supplies to the company’s processing facilities (factories, etc.).
The next kind of application in Table 6.1 (coordinating product mixes at plants) illustrates that arcs can represent something other than a shipping lane for a physical flow of
materials. This application involves a company with several plants (the supply nodes)
that can produce the same products but at different costs. Each arc from a supply node
represents the production of one of the possible products at that plant, where this arc
leads to the transshipment node that corresponds to this product. Thus, this transshipment
node has an arc coming in from each plant capable of producing this product, and then
the arcs leading out of this node go to the respective customers (the demand nodes) for
this product. The objective is to determine how to divide each plant’s production capacity
among the products so as to minimize the total cost of meeting the demand for the various
products.
The last application in Table 6.1 (cash flow management) illustrates that different
nodes can represent some event that occurs at different times. In this case, each supply
node represents a specific time (or time period) when some cash will become available to
the company (through maturing accounts, notes receivable, sales of securities, borrowing,
etc.). The supply at each of these nodes is the amount of cash that will become available
then. Similarly, each demand node represents a specific time (or time period) when the
company will need to draw on its cash reserves. The demand at each such node is the
amount of cash that will be needed then. The objective is to maximize the company’s
income from investing the cash between each time it becomes available and when it will
be used. Therefore, each transshipment node represents the choice of a specific short-term
investment option (e.g., purchasing a certificate of deposit from a bank) over a specific
time interval. The resulting network will have a succession of flows representing a schedule for cash becoming available, being invested, and then being used after the maturing of
the investment.
Special Types of Minimum-Cost Flow Problems
transshipment problem A
transshipment problem is
just a minimum-cost flow
problem that has unlimited
capacities for all its arcs.
There are five important categories of network problems that turn out to be special types of
minimum-cost flow problems.
One is the transportation problems discussed in Section 3.5.Recall that a typical kind of
transportation problem involves minimizing the cost of shipping something (e.g., many units
of a particular product) from a number of sources (e.g., plants producing this product) directly
to a number of destinations (e.g., customers). Figure 3.9 shows the network representation of
a typical transportation problem. In our current terminology, the sources and destinations of
a transportation problem are the supply nodes and demand nodes, respectively. Thus, a transportation problem is just a minimum-cost flow problem without any transshipment nodes and
without any capacity constraints on the arcs (all of which go directly from a supply node to a
demand node).
A second category is the assignment problems discussed in Section 3.6. Recall that
this kind of problem involves assigning a group of people (or other operational units) to a
group of tasks where each person is to perform a single task. An assignment problem can
be viewed as a special type of transportation problem whose sources are the assignees and
whose destinations are the tasks. This then makes the assignment problem also a special type
of minimum-cost flow problem with the characteristics described in the preceding paragraph.
In addition, each person is a supply node with a supply of 1 and each task is a demand node
with a demand of 1.
A third special type of minimum-cost flow problem is transshipment problems. This
kind of problem is just like a transportation problem except for the additional feature that
the shipments from the sources (supply nodes) to the destinations (demand nodes) might
also pass through intermediate transfer points (transshipment nodes) such as distribution
centers. Like a transportation problem, there are no capacity constraints on the arcs. Consequently, any minimum-cost flow problem where each arc can carry any desired amount of
212 Chapter Six Network Optimization Problems
The network simplex
method can be used to solve
huge problems of any of
these five special types.
Review
Questions
6.2
flow is a transshipment problem. For example, if the data in Figure 6.2 were altered so that
any amounts (within the ranges of the supplies and demands) could be shipped into and out
of the distribution center, the Distribution Unlimited Co. would become just a transshipment problem.1
Because of their close relationship to a general minimum-cost flow problem, we will not
discuss transshipment problems further.
The other two important special types of minimum-cost flow problems are maximum
flow problems and shortest path problems, which will be described in Sections 6.3 and 6.4
after presenting a case study of a maximum flow problem in the next section.
In case you are wondering why we are bothering to point out that these five kinds of
problems are special types of minimum-cost flow problems, here is one very important reason. It means that the network simplex method can be used to solve huge problems of any of
these types that might be difficult or impossible for the simplex method to solve. It is true
that other efficient special-purpose algorithms also are available for each of these kinds of
problems. However, recent implementations of the network simplex method have become so
powerful that it now provides an excellent alternative to these other algorithms in most cases.
This is especially valuable when the available software package includes the network simplex
method but not another relevant special-purpose algorithm. Furthermore, even after finding
an optimal solution, the network simplex method can continue to be helpful in aiding managerial what-if sessions along the lines discussed in Chapter 5.
1.
2.
3.
4.
5.
6.
Name and describe the three kinds of nodes in a minimum-cost flow problem.
What is meant by the capacity of an arc?
What is the usual objective for a minimum-cost flow problem?
What property is necessary for a minimum-cost flow problem to have feasible solutions?
What is the integer solutions property for minimum-cost flow problems?
What is the name of the streamlined version of the simplex method that is designed to
solve minimum-cost flow problems very efficiently?
7. What are a few typical kinds of applications of minimum-cost flow problems?
8. Name five important categories of network optimization problems that turn out to be s­ pecial
types of minimum-cost flow problems.
A CASE STUDY: THE BMZ CO. MAXIMUM FLOW PROBLEM
What a day! First being called into his boss’s office and then receiving an urgent telephone
call from the company president himself. Fortunately, he was able to reassure them that he has
the situation under control.
Although his official title is Supply Chain Manager for the BMZ Company, Karl Schmidt
often tells his friends that he really is the company’s crisis manager. One crisis after another.
The supplies needed to keep the production lines going haven’t arrived yet. Or the supplies
have arrived but are unusable because they are the wrong size. Or an urgent shipment to a
key customer has been delayed. This current crisis is typical. One of the company’s most
important distribution centers—the one in Los Angeles—urgently needs an increased flow of
shipments from the company.
Karl was chosen for this key position because he is considered a rising young star. ­Having
just received his MBA degree from a top American business school four years ago, he is
the youngest member of upper-level management in the entire company. His business school
training in the latest management science techniques has proven invaluable in improving supply chain management throughout the company. The crises still occur, but the frequent chaos
of past years has been eliminated.
1
Be aware that a minimum-cost flow problem that does have capacity constraints on the arcs is sometimes
referred to as a capacitated transshipment problem. We will not use this terminology.
6.2 A Case Study: The Bmz Co. Maximum Flow Problem 213
Karl has a plan for dealing with the current crisis. This will mean calling on management
science once again.
Background
The BMZ Company is a European manufacturer of luxury automobiles. Although its cars
sell well in all the developed countries, its exports to the United States are particularly important to the company.
BMZ has a well-deserved reputation for providing excellent service. One key to maintaining this reputation is having a plentiful supply of automobile replacement parts readily
available to the company’s numerous dealerships and authorized repair shops. These parts are
mainly stored in the company’s distribution centers and then delivered promptly when needed.
One of Karl Schmidt’s top priorities is avoiding shortages at these distribution centers.
The company has several distribution centers in the United States. However, the closest
one to the Los Angeles center is over 1,000 miles away in Seattle. Since BMZ cars are becoming especially popular in California, it is particularly important to keep the Los Angeles center
well supplied. Therefore, the fact that supplies there are currently dwindling is a matter of real
concern to BMZ top management—as Karl learned forcefully today.
Most of the automobile replacement parts are produced at the company’s main factory in
Stuttgart, Germany, along with the production of new cars. It is this factory that has been supplying the Los Angeles center with spare parts. Some of these parts are bulky, and very large
numbers of certain parts are needed, so the total volume of the supplies has been relatively
massive—over 300,000 cubic feet of goods arriving monthly. Now a much larger amount will
be needed over the next month to replenish the dwindling inventory.
The Problem
The problem is to maximize
the flow of automobile
replacement parts from the
factory in Stuttgart, Germany, to the distribution
center in Los Angeles.
Karl needs to execute a plan quickly for shipping as much as possible from the main factory
to the distribution center in Los Angeles over the next month. He already has recognized that
this is a maximum flow problem—a problem of maximizing the flow of replacement parts
from the factory to this distribution center.
The factory is producing far more than can be shipped along a single route to this one
distribution center. Therefore, the limiting factor on how much can be shipped is the limited
capacity of the company’s distribution network.
This distribution network is depicted in Figure 6.6, where the nodes labeled ST and LA are
the factory in Stuttgart and the distribution center in Los Angeles, respectively. There is a rail
FIGURE 6.6
The BMZ Co. distribution
network from its main
factory in Stuttgart,
­Germany, to a distribution
center in Los Angeles.
RO
New York
0
[8
LA
Los Angeles
ts
i
un
ts
uni
NY
[60
[40 units max.]
x.]
.]
ax
m
its
New Orleans
[70 units
max.]
[50
x.]
ma
a
m
un
0
[5
ax.]
ts m
i
n
30 u
NO
[
Rotterdam
un
its
BO
ma
x.]
[70 units max.]
x.]
Bordeaux
ma
s
t
i
un
[40
LI
Lisbon
ST
Stuttgart
214 Chapter Six Network Optimization Problems
head at the factory, so shipments first go by rail to one of three European ports: ­Rotterdam
(node RO), Bordeaux (node BO), and Lisbon (node LI). They then go by ship to ports in
the United States, either New York (node NY) or New Orleans (node NO). Finally, they are
shipped by truck from these ports to the distribution center in Los Angeles.
The organizations operating these railroads, ships, and trucks are independently owned
companies that ship goods for numerous firms. Because of prior commitments to their regular customers, these companies are unable to drastically increase the allocation of space to
any single customer on short notice. Therefore, the BMZ Co. is only able to secure a limited
amount of shipping space along each shipping lane over the next month. The amounts available
are given in Figure 6.6, using units of hundreds of cubic meters. (Since each unit of 100 cubic
meters is a little over 3,500 cubic feet, these are large volumes of goods that need to be moved.)
Model Formulation
In contrast to the spreadsheet model in Figure 6.5,
which minimizes TotalCost
(D11), the spreadsheet
model in Figure 6.8
­maximizes the objective
cell MaxFlow (D14).
Figure 6.7 shows the network model for this maximum flow problem. Rather than showing the
geographical layout of the distribution network, this network simply lines up the nodes (representing the cities) in evenly spaced columns. The arcs represent the shipping lanes, where
the capacity of each arc (given in square brackets under the arc) is the amount of shipping
space available along that shipping lane. The objective is to determine how much flow to send
through each arc (how many units to ship through each shipping lane) to maximize the total
number of units flowing from the factory in Stuttgart to the distribution center in Los Angeles.
Figure 6.8 shows the corresponding spreadsheet model for this problem when using the
format introduced in Figure 6.5. The main difference from the model in Figure 6.5 is the
change in the objective. Since we are no longer minimizing the total cost of the flow through
the network, column G in Figure 6.5 can be deleted in Figure 6.8. The objective cell MaxFlow (D14) in Figure 6.8 now needs to give the total number of units flowing from Stuttgart
to Los Angeles. Thus, the equations at the bottom of the figure include D14 = I4, where I4
gives the net flow leaving Stuttgart to go to Los Angeles. As in Figure 6.5, the equations in
Figure 6.8 entered into NetFlow (I4:I10) again use the difference of two SUMIF functions
to calculate the net flow generated at each node. Since the objective is to maximize the flow
shown in MaxFlow (D14), the Solver Parameters box specifies that this objective cell is to
be ­maximized. After running Solver, the optimal solution shown in the changing cells Ship
(D4:D12) is obtained for the amount that BMZ should ship through each shipping lane.
FIGURE 6.7
RO
[6
0]
A network model for the
BMZ Co. problem as a
maximum flow problem,
where the number in
square brackets below
each arc is the capacity of
that arc.
[50
0]
[8
[4
0]
]
NY
BO
[70]
]
0
[7
[5
0]
LA
[40
]
NO
]
0
[3
LI
ST
6.3 Maximum Flow Problems 215
FIGURE 6.8
A spreadsheet model for the BMZ Co. maximum flow problem, including the equations entered into the objective cell MaxFlow
(D14) and the other output cells NetFlow (I4:I10), as well as the other specifications needed to set up the model. The changing cells
Ship (D4:D12) show the optimal shipping quantities through the distribution network obtained by Solver.
A
1
B
C
D
E
F
G
J
K
H
I
Nodes
Net Flow
50
Stuttgart
150
70
Rotterdam
0
=
0
Bordeaux
0
=
0
BMZ Co. Maximum Flow Problem
2
To
From
3
Ship
Capacity
≤
4
Stuttgart
Rotterdam
50
5
Stuttgart
Bordeaux
70
≤
6
Stuttgart
Lisbon
30
≤
40
Supply/Demand
7
Rotterdam
New York
50
≤
60
Lisbon
0
=
0
8
Bordeaux
New York
30
≤
40
New York
0
0
9
Bordeaux
New Orleans
40
≤
=
50
New Orleans
0
=
0
30
Los Angeles
-150
10
Lisbon
New Orleans
30
≤
11
New York
Los Angeles
80
≤
80
12
New Orleans
Los Angeles
70
≤
70
13
Maximum Flow
14
Solver Parameters
Set Objective Cell: MaxFlow
To: Max
By Changing Variable Cells:
Ship
Subject to the Constraints:
I5:I9 = SupplyDemand
Ship <= Capacity
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
150
Range Name
Cells
Capacity
From
MaxFlow
NetFlow
Nodes
Ship
SupplyDemand
To
F4:F12
B4:B12
D14
I4:I10
H4:H10
D4:D12
K5:K9
C4:C12
I
Net Flow
3
4
=SUMIF(From,H4,Ship)-SUMIF(To,H4,Ship)
5
=SUMIF(From,H5,Ship)-SUMIF(To,H5,Ship)
6
=SUMIF(From,H6,Ship)-SUMIF(To,H6,Ship)
7
=SUMIF(From,H7,Ship)-SUMIF(To,H7,Ship)
8
=SUMIF(From,H8,Ship)-SUMIF(To,H8,Ship)
9
=SUMIF(From,H9,Ship)-SUMIF(To,H9,Ship)
10
=SUMIF(From,H10,Ship)-SUMIF(To,H10,Ship)
14
C
D
Maximum Flow
=I4
However, Karl is not completely satisfied with this solution. He has an idea for doing even
better. This will require formulating and solving another maximum flow problem. (This story
continues in the middle of the next section.)
Review
Questions
6.3
1. What is the current crisis facing the BMZ Co.?
2. When formulating this problem in network terms, what is flowing through BMZ’s distribution network? From where to where?
3. What is the objective of the resulting maximum flow problem?
MAXIMUM FLOW PROBLEMS
Like a minimum-cost flow problem, a maximum flow problem is concerned with flow through
a network. However, the objective now is different. Rather than minimizing the cost of the
flow, the objective now is to find a flow plan that maximizes the amount flowing through the
network. This is how Karl Schmidt was able to find a flow plan that maximizes the number of
units of automobile replacement parts flowing through BMZ’s distribution network from its
factory in Stuttgart to the distribution center in Los Angeles.
An Application Vignette
The network for transport of natural gas on the Norwegian Continental Shelf, with approximately 5,000 miles of subsea pipelines, is the world’s largest offshore pipeline network. Gassco is
a company entirely owned by the Norwegian state, which operates this network. Another company that is largely state owned,
StatoilHydro, is the main Norwegian supplier of natural gas to
markets throughout Europe and elsewhere.
Gassco and StatoilHydro together use management science
techniques to optimize both the configuration of the network
and the routing of the natural gas. The main model used for this
routing is a multicommodity network-flow model in which the
different hydrocarbons and contaminants in natural gas constitute the commodities. The objective function for the model is to
maximize the total flow of the natural gas from the supply points
(the offshore drilling platforms) to the demand points (typically
import terminals). However, in addition to the usual supply-anddemand constraints, the model also includes constraints involving pressure-flow relationships, maximum delivery pressures,
and technical pressure bounds on pipelines. Therefore, this
model is a generalization of the model for the maximum flow
problem described in this section.
This key application of management science, along with a
few others, has had a dramatic impact on the efficiency of the
operation of this offshore pipeline network. The resulting accumulated savings were estimated to be approximately $2 billion
in the period 1995–2008.
Source: F. Romo, A. Tomasgard, L. Hellemo, M. Fodstad, B. H. Eidesen,
and B. Pedersen, “Optimizing the Norwegian Natural Gas Production
and Transport,” Interfaces 39, no. 1 (January–February 2009), pp. 46–56.
(A link to this article is provided at www.mhhe.com/Hillier6e.)
General Characteristics
Except for the difference in objective (maximize flow versus minimize cost), the characteristics of the maximum flow problem are quite similar to those for the minimum-cost flow
problem. However, there are some minor differences, as we will discuss after summarizing
the assumptions.
Assumptions of a Maximum Flow Problem
The objective is to find a
flow plan that maximizes
the flow from the source to
the sink.
Although a maximum flow
problem has only a single
source and a single sink,
variants with multiple
sources and sinks also can
be solved, as illustrated in
the next subsection.
216
1. All flow through the network originates at one node, called the source, and terminates
at one other node, called the sink. (The source and sink in the BMZ problem are the factory in Stuttgart and the distribution center in Los Angeles, respectively, as represented by
nodes ST and LA in Figure 6.7.)
2. All the remaining nodes are transshipment nodes. (These are nodes RO, BO, LI, NY, and
NO in the BMZ problem.)
3. Flow through an arc is only allowed in the direction indicated by the arrowhead, where the
maximum amount of flow is given by the capacity of that arc. At the source, all arcs point
away from the node. At the sink, all arcs point into the node.
4. The objective is to maximize the total amount of flow from the source to the sink. This
amount is measured in either of two equivalent ways, namely, either the amount leaving
the source or the amount entering the sink. (Cells D14 and I4 in Figure 6.8 use the amount
leaving the source.)
The source and sink of a maximum flow problem are analogous to the supply nodes and
demand nodes of a minimum-cost flow problem. These are the only nodes in both problems
that do not have conservation of flow (flow out equals flow in). Like the supply nodes, the
source generates flow. Like the demand nodes, the sink absorbs flow.
However, there are two differences between these nodes in a minimum-cost flow problem
and the corresponding nodes in a maximum flow problem.
One difference is that, whereas supply nodes have fixed supplies and demand nodes have
fixed demands, the source and sink do not. The reason is that the objective is to maximize the
flow leaving the source and entering the sink rather than fixing this amount.
The second difference is that, whereas the number of supply nodes and the number of
demand nodes in a minimum-cost flow problem may be more than one, there can be only one
source and only one sink in a maximum flow problem. However, variants of maximum flow
problems that have multiple sources and sinks can still be solved by Solver, as you now will
see illustrated by the BMZ case study introduced in the preceding section.
6.3 Maximum Flow Problems 217
Continuing the Case Study with Multiple Supply Points and Multiple
Demand Points
Here is Karl Schmidt’s idea for how to improve upon the flow plan obtained at the end of
­Section 6.2 (as given in column D of Figure 6.8).
The company has a second, smaller factory in Berlin, north of its Stuttgart factory, for producing automobile parts. Although this factory normally is used to help supply distribution
centers in northern Europe, Canada, and the northern United States (including one in Seattle),
it also is able to ship to the distribution center in Los Angeles. Furthermore, the distribution
center in Seattle has the capability of supplying parts to the customers of the distribution center in Los Angeles when shortages occur at the latter center.
In this light, Karl now has developed a better plan for addressing the current inventory
shortages in Los Angeles. Rather than simply maximizing shipments from the Stuttgart factory to Los Angeles, he has decided to maximize the total shipments from both factories to the
distribution centers in both Los Angeles and Seattle.
Figure 6.9 shows the network model representing the expanded distribution network that
encompasses both factories and both distribution centers. In addition to the nodes shown in
Figures 6.6 and 6.7, node BE is the second, smaller factory in Berlin; nodes HA and BN are
additional ports used by this factory in Hamburg and Boston, respectively; and node SE is the
distribution center in Seattle. As before, the arcs represent the shipping lanes, where the number in square brackets below each arc is the capacity of that arc, that is, the maximum number
of units that can be shipped through that shipping lane over the next month.
The corresponding spreadsheet model is displayed in Figure 6.10. The format is the same
as in Figure 6.8. However, the objective cell MaxFlow (D21) now gives the total flow from
Stuttgart and Berlin, so D21 = I4 + I5 (as shown by the equation for this objective cell given
at the bottom of the figure).
The changing cells Ship (D4:D19) in this figure show the optimal solution obtained for
the number of units to ship through each shipping lane over the next month. Comparing this
solution with the one in Figure 6.8 shows the impact of Karl Schmidt’s decision to expand the
distribution network to include the second factory and the distribution center in Seattle. As
indicated in column I of the two figures, the number of units going to Los Angeles directly
has been increased from 150 to 160, in addition to the 60 units going to Seattle as a backup
for the inventory shortage in Los Angeles. This plan solved the crisis in Los Angeles and won
Karl commendations from top management.
FIGURE 6.9
HA
BN
]
[40
[6
0]
]
[30
]
[20
SE
RO
[40
]
]
0]
0]
[5
]
LA
BE
[20]
[60
NY
[1
A network model for the
expanded BMZ Co. problem as a variant of a maximum flow problem, where
the number in square
brackets below each arc is
the capacity of that arc.
[40
[80
[70
]
]
BO
ST
[70]
]
[50
0]
NO
[4
[30
]
LI
218 Chapter Six Network Optimization Problems
FIGURE 6.10
A spreadsheet model for the expanded BMZ Co. problem as a variant of a maximum flow problem with sources in both Stuttgart and
Berlin and sinks in both Los Angeles and Seattle. Using the objective cell MaxFlow (D21) to maximize the total flow from the two
sources to the two sinks, Solver yields the optimal shipping plan shown in the changing cells Ship (D4:D19).
A
1
B
C
D
E
F
G
J
H
I
Capacity
Nodes
Net Flow
K
BMZ Co. Expanded Maximum Flow Problem
2
Supply/Demand
3
From
To
Ship
4
Stuttgart
Rotterdam
40
≤
50
Stuttgart
140
5
Stuttgart
Bordeaux
70
≤
70
Berlin
80
6
Stuttgart
Lisbon
30
≤
40
Hamburg
0
=
0
7
Berlin
Rotterdam
20
≤
20
Rotterdam
0
=
0
8
Berlin
Hamburg
60
≤
60
Bordeaux
0
=
0
9
Rotterdam
New York
60
≤
60
Lisbon
0
=
0
≤
40
Boston
0
=
0
10
Bordeaux
New York
30
11
Bordeaux
New Orleans
40
≤
50
New York
0
=
0
12
Lisbon
New Orleans
30
≤
30
New Orleans
0
=
0
13
Hamburg
New York
30
≤
30
Los Angeles
-160
14
Hamburg
Boston
30
≤
40
Seattle
-60
15
New Orleans
Los Angeles
70
≤
70
16
New York
Los Angeles
80
≤
80
17
New York
Seattle
40
≤
40
18
Boston
Los Angeles
10
≤
10
20
≤
20
19
Boston
Seattle
20
21
Maximum Flow
Solver Parameters
Set Objective Cell: MaxFlow
To: Max
By Changing Variable Cells:
Ship
Subject to the Constraints:
I6:I12 = SupplyDemand
Ship <= Capacity
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
220
Range Name
Cells
Capacity
From
MaxFlow
NetFlow
Nodes
Ship
SupplyDemand
To
F4:F19
B4:B19
D21
I4:I14
H4:H14
D4:D19
K6:K12
C4:C19
I
3
Net Flow
4
=SUMIF(From,H4,Ship)-SUMIF(To,H4,Ship)
5
=SUMIF(From,H5,Ship)-SUMIF(To,H5,Ship)
6
=SUMIF(From,H6,Ship)-SUMIF(To,H6,Ship)
7
=SUMIF(From,H7,Ship)-SUMIF(To,H7,Ship)
8
=SUMIF(From,H8,Ship)-SUMIF(To,H8,Ship)
9
=SUMIF(From,H9,Ship)-SUMIF(To,H9,Ship)
10
=SUMIF(From,H10,Ship)-SUMIF(To,H10,Ship)
11
=SUMIF(From,H11,Ship)-SUMIF(To,H11,Ship)
12
=SUMIF(From,H12,Ship)-SUMIF(To,H12,Ship)
13
=SUMIF(From,H13,Ship)-SUMIF(To,H13,Ship)
14
=SUMIF(From,H14,Ship)-SUMIF(To,H14,Ship)
21
C
D
Maximum Flow
=I4+I5
Some Applications
The applications of maximum flow problems and their variants are somewhat similar to those
for minimum-cost flow problems described in the preceding section when management’s objective is to maximize flow rather than to minimize cost. Here are some typical kinds of applications.
1. Maximize the flow through a distribution network, as for the BMZ Co. problem.
2. Maximize the flow through a company’s supply network from its vendors to its processing
facilities.
6.4 Shortest Path Problems 219
3. Maximize the flow of oil through a system of pipelines.
4. Maximize the flow of water through a system of aqueducts.
5. Maximize the flow of vehicles through a transportation network.
Solving Very Large Problems
The expanded BMZ network in Figure 6.9 has 11 nodes and 16 arcs. However, the networks
for most real applications are considerably larger, and occasionally vastly larger with many
thousands of nodes and arcs. For huge networks, Solver may not be capable of solving such
large maximum flow problems even if they still can be formulated as illustrated in Figures 6.8
and 6.10 .
Fortunately, management scientists have other ways of efficiently formulating huge
maximum flow problems and then solving them. For example, an extremely efficient
­special-purpose algorithm has been developed specifically for solving maximum flow
problems. Another solution approach is to reformulate the problem to fit the format for
a minimum-cost flow problem so that the network simplex method can be applied. These
special algorithms are available in some software packages, but not in Solver. Thus, if you
should ever encounter a maximum flow problem or a variant that is beyond the scope of
Solver (which won’t happen in this book), rest assured that it probably can be formulated
and solved in another way.
Review
Questions
6.4
1. How does the objective of a maximum flow problem differ from that for a minimum-cost
flow problem?
2. What are the source and the sink for a maximum flow problem? For each, in what direction
do all their arcs point?
3. What are the two equivalent ways in which the total amount of flow from the source to the
sink can be measured?
4. The source and sink of a maximum flow problem are different from the supply nodes and
demand nodes of a minimum-cost flow problem in what two ways?
5. What are a few typical kinds of applications of maximum flow problems?
SHORTEST PATH PROBLEMS
The most common applications of shortest path problems are for what the name suggests—
finding the shortest path between two points. Here is an example.
An Example: The Littletown Fire Department Problem
The objective is to find
the shortest route from the
fire station to the farming
community.
Littletown is a small town in a rural area. Its fire department serves a relatively large geographical area that includes many farming communities. Since there are numerous roads
throughout the area, many possible routes may be available for traveling to any given farming
community from the fire station. Since time is of the essence in reaching a fire, the fire chief
wishes to determine in advance the shortest path from the fire station to each of the farming
communities.
Figure 6.11 shows the road system connecting the fire station to one of the farming communities, including the mileage along each road. Can you find which route from the fire station to the farming community minimizes the total number of miles?
Model Formulation for the Littletown Problem
Figure 6.12 gives a network representation of this problem, which ignores the geographical
­layout and the curves in the roads. (When lining up the nodes in columns as in this figure, the
selection of which nodes should go together in a column is somewhat arbitrary, but the ­selections
here fit both the geography and the road system well.) This network model is the usual way
of representing a shortest path problem. The junctions now are nodes of the network, where
the fire station and farming community are two additional nodes labeled as O (for origin) and T
An Application Vignette
Incorporated in 1881, Canadian Pacific Railway (CPR) was
North America’s first transcontinental railway. CPR transports
rail freight over a 14,000-mile network extending from Montreal
to Vancouver and throughout the U.S. Northwest and Midwest.
Alliances with other carriers extend CPR’s market reach into the
major business centers of Mexico as well.
Every day CPR receives approximately 7,000 new shipments
from its customers going to destinations across North America
and for export. It must route and move these shipments in railcars over the network of track, where a railcar may be switched
a number of times from one locomotive engine to another before
reaching its destination. CPR must coordinate the shipments with
its operational plans for approximately 1,600 locomotives, 65,000
railcars, over 5,000 train crew members, and 250 train yards.
CPR management turned to a management science consulting firm, MultiModal Applied Systems, to work with CPR
employees in developing a management science approach
to this problem. A variety of management science techniques
were used to create a new operating strategy. However, the
foundation of the approach was to represent the flow of blocks
links In a shortest path
problem, travel goes from
the origin to the destination
through a series of links
(such as roads) that connect
pairs of nodes (junctions) in
the network.
of railcars as flow through a network where each node corresponds to both a location and a point in time. This representation then enabled the application of network optimization
techniques. For example, numerous shortest path problems are
solved each day as part of the overall approach.
This application of management science is saving CPR
roughly US$100 million per year. Labor productivity, locomotive productivity, fuel consumption, and railcar velocity have
improved very substantially. In addition, CPR now provides its
customers with reliable delivery times and has received many
awards for its improvement in service. This application of network optimization techniques also led to CPR winning the prestigious First Prize in the 2003 international competition for the
Franz Edelman Award for Achievement in Operations Research
and the Management Sciences.
Source: P. Ireland, R. Case, J. Fallis, C. Van Dyke, J. Kuehn, and M.
Meketon, “The Canadian Pacific Railway Transforms Operations by
Using Models to Develop Its Operating Plans,” Interfaces 34, no. 1
(January–February 2004), pp. 5–14. (A link to this article is provided
at www.mhhe.com/Hillier6e.)
(for destination), respectively. Since travel (flow) can go in either direction between the nodes,
the lines connecting the nodes now are referred to as links2 instead of arcs. A link between a
pair of nodes allows travel in either direction, whereas an arc allows travel in only the direction
indicated by an arrowhead, so the lines in Figure 6.12 need to be links instead of arcs. (Notice
that the links do not have an arrowhead at either end.)
8
FIGURE 6.11
The road system between
the Littletown Fire Station
and a certain farming community, where A, B, . . .,
H are junctions and the
number next to each road
shows its distance in
miles.
6
A
1
3
6
Fire
Station
D
4
6
3
(Origin)
O
1
6
4
2
C
2
220
6
5
7
3
H
4
3
6
5
E
2
F
8
D
4
B
4
7
A
The network representation of Figure 6.11 as a
shortest path problem.
5
E
2
C
Farming
Community
G
5
FIGURE 6.12
6
3
3
B
4
4
F
4
G
2
6
T
(Destination)
7
H
Another name sometimes used is undirected arc, but we will not use this terminology.
7
6.4 Shortest Path Problems 221
Have you found the shortest path from the origin to the destination yet? (Try it now before
reading further.) It is
​O → A → B → E → F → T​
This spreadsheet model is
like one for a minimum-cost
flow problem with no arc
capacity constraints except
that distances replace unit
costs and travel on a chosen
path is interpreted as a flow
of 1 through this path.
with a total distance of 19 miles.
This problem (like any shortest path problem) can be thought of as a special kind of
minimum-cost flow problem (Section 6.1) where the miles traveled now are interpreted to be
the cost of flow through the network. A trip from the fire station to the farming community is
interpreted to be a flow of 1 on the chosen path through the network, so minimizing the cost of
this flow is equivalent to minimizing the number of miles traveled. The fire station is considered
to be the one supply node, with a supply of 1 to represent the start of this trip. The farming community is the one demand node, with a demand of 1 to represent the completion of this trip. All
the other nodes in Figure 6.12 are transshipment nodes, so the net flow generated at each is 0.
Figure 6.13 shows the spreadsheet model that results from this interpretation. The format
is basically the same as for the minimum-cost flow problem formulated in Figure 6.5, except
now there are no arc capacity constraints and the unit cost column is replaced by a column
of distances in miles. The flow quantities given by the changing cells OnRoute (D4:D27) are
1 for each arc that is on the chosen path from the fire station to the farming community and
0 otherwise. The objective cell TotalDistance (D29) gives the total distance of this path in
miles. (See the equation for this cell at the bottom of the figure.) Columns B and C together
list all the vertical links in Figure 6.12 twice, once as a downward arc and once as an upward
arc, since either direction might be on the chosen path. The other links are only listed as leftto-right arcs, since this is the only direction of interest for choosing a shortest path from the
origin to the destination.
Column K shows the net flow that needs to be generated at each of the nodes. Using the
equations at the bottom of the figure, each column I cell then calculates the actual net flow at
that node by adding the flow out and subtracting the flow in. The corresponding constraints,
Nodes (H4:H13) = SupplyDemand (K4:K13), are specified in the Solver Parameters box.
The solution shown in OnRoute (D4:D27) is the optimal solution obtained after running
Solver. It is exactly the same as the shortest path given earlier.
Just as for minimum-cost flow problems and maximum flow problems, special algorithms
are available for solving large shortest path problems very efficiently, but these algorithms are
not included in Solver. Using a spreadsheet formulation and Solver is fine for problems of the
size of the Littletown problem and somewhat larger, but you should be aware that vastly larger
problems can still be solved by other means.
General Characteristics
Except for more complicated variations beyond the scope of this book, all shortest path problems
share the characteristics illustrated by the Littletown problem. Here are the basic assumptions.
Assumptions of a Shortest Path Problem
The objective is to find
the shortest path from the
­origin to the destination.
1. You need to choose a path through the network that starts at a certain node, called the
­origin, and ends at another certain node, called the destination.
2. The lines connecting certain pairs of nodes commonly are links (which allow travel in
either direction), although arcs (which only permit travel in one direction) also are allowed.
3. Associated with each link (or arc) is a nonnegative number called its length. (Be aware
that the drawing of each link in the network typically makes no effort to show its true
length other than giving the correct number next to the link.)
4. The objective is to find the shortest path (the path with the minimum total length) from the
origin to the destination.
Some Applications
Not all applications of shortest path problems involve minimizing the distance traveled from
the origin to the destination. In fact, they might not even involve travel at all. The links (or
arcs) might instead represent activities of some other kind, so choosing a path through the
222 Chapter Six Network Optimization Problems
FIGURE 6.13
A spreadsheet model for the Littletown Fire Department shortest path problem, including the equations entered into the objective cell
TotalDistance (D29) and the other output cells SupplyDemand (K4:K13). The values of 1 in the changing cells OnRoute (D4:D27)
reveal the optimal solution obtained by Solver for the shortest path (19 miles) from the fire station to the farming community.
A
1
B
C
D
E
F
G
H
I
J
K
Littletown Fire Department Shortest Path Problem
2
3
From
To
On Route
Distance
Nodes
Net Flow
4
Supply/Demand
Fire St.
A
1
3
Fire St.
1
=
1
5
Fire St.
B
0
6
A
0
=
0
6
Fire St.
C
0
4
B
0
=
0
7
A
B
1
1
C
0
=
0
8
A
D
0
6
D
0
=
0
9
B
A
0
1
E
0
=
0
10
B
C
0
2
F
0
=
0
11
B
D
0
4
G
0
=
0
12
B
E
1
5
H
0
=
0
13
C
B
0
2
Farm Comm.
–1
=
–1
14
C
E
0
7
15
D
E
0
3
16
D
F
0
8
17
E
D
0
3
18
E
F
1
6
19
E
G
0
5
20
E
H
0
4
21
F
G
0
3
22
F
Farm Comm.
1
4
23
G
F
0
3
24
G
H
0
2
25
G
Farm Comm.
0
6
26
H
G
0
2
27
H
Farm Comm.
0
7
Total Distance
19
28
29
Solver Parameters
I
Set Objective Cell: TotalDistance
To: Min
By Changing Variable Cells:
OnRoute
Subject to the Constraints:
NetFlow = SupplyDemand
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
Range Name
Cells
Distance
From
NetFlow
Nodes
OnRoute
SupplyDemand
To
TotalDistance
F4:F27
B4:B27
I4:I13
H4:H13
D4:D27
K4:K13
C4:C27
D29
3
Net Flow
4
=SUMIF(From,H4,OnRoute)-SUMIF(To,H4,OnRoute)
5
=SUMIF(From,H5,OnRoute)-SUMIF(To,H5,OnRoute)
6
=SUMIF(From,H6,OnRoute)-SUMIF(To,H6,OnRoute)
7
=SUMIF(From,H7,OnRoute)-SUMIF(To,H7,OnRoute)
8
=SUMIF(From,H8,OnRoute)-SUMIF(To,H8,OnRoute)
9
=SUMIF(From,H9,OnRoute)-SUMIF(To,H9,OnRoute)
10
=SUMIF(From,H10,OnRoute)-SUMIF(To,H10,OnRoute)
11
=SUMIF(From,H11,OnRoute)-SUMIF(To,H11,OnRoute)
12
=SUMIF(From,H12,OnRoute)-SUMIF(To,H12,OnRoute)
13
=SUMIF(From,H13,OnRoute)-SUMIF(To,H13,OnRoute)
29
C
D
Total Distance
=SUMPRODUCT(OnRoute,Distance)
6.4 Shortest Path Problems 223
network corresponds to selecting the best sequence of activities. The numbers giving the
“lengths” of the links might then be, for example, the costs of the activities, in which case the
objective would be to determine which sequence of activities minimizes the total cost.
Here are three categories of applications.
1. Minimize the total distance traveled, as in the Littletown example.
2. Minimize the total cost of a sequence of activities, as in the example that follows in the
subsection below.
3. Minimize the total time of a sequence of activities, as in the example involving the Quick
Company at the end of this section.
An Example of Minimizing Total Cost
Sarah needs a schedule for
trading in her car that will
minimize her total net cost.
Sarah has just graduated from high school. As a graduation present, her parents have given her
a car fund of $21,000 to help purchase and maintain a certain three-year-old used car for college.
Since operating and maintenance costs go up rapidly as the car ages, Sarah’s parents tell her that
she will be welcome to trade in her car on another three-year-old car one or more times during
the next three summers if she determines that this would minimize her total net cost. They also
inform her that they will give her a new car in four years as a college graduation present, so she
should definitely plan to trade in her car then. (These are pretty nice parents!)
Table 6.2 gives the relevant data for each time Sarah purchases a three-year-old car. For
example, if she trades in her car after two years, the next car will be in ownership year 1 during her junior year, and so forth.
When should Sarah trade in her car (if at all) during the next three summers to minimize her
total net cost of purchasing, operating, and maintaining the car(s) over her four years of college?
Figure 6.14 shows the network formulation of this problem as a shortest path problem.
Nodes 1, 2, 3, and 4 are the end of Sarah’s first, second, third, and fourth years of college,
respectively. Node 0 is now, before starting college. Each arc from one node to a second node
corresponds to the activity of purchasing a car at the time indicated by the first of these two
nodes and then trading it in at the time indicated by the second node. Sarah begins by purchasing a car now, and she ends by trading in a car at the end of year 4, so node 0 is the origin and
node 4 is the destination.
The number of arcs on the path chosen from the origin to the destination indicates how
many times Sarah will purchase and trade in a car. For example, consider the path
0
TABLE 6.2
Sarah’s Data Each Time
She Purchases a ThreeYear-Old Car
1
4
Operating and Maintenance Costs
for Ownership Year
Purchase
Price
$12,000
Trade-in Value at End
of Ownership Year
1
2
3
4
1
2
3
4
$2,000
$3,000
$4,500
$6,500
$8,500
$6,500
$4,500
$3,000
25,000
FIGURE 6.14
Formulation of the
problem of when Sarah
should trade in her car as
a shortest path problem.
The node labels measure
the number of years from
now. Each arc represents
purchasing a car and then
trading it in later.
3
17,000
10,500
10,500
(Origin)
0
5,500
1
5,500
2
5,500
10,500
17,000
3
5,500
4
(Destination)
224 Chapter Six Network Optimization Problems
This corresponds to purchasing a car now, then trading it in at the end of year 1 to purchase a
second car, then trading in the second car at the end of year 3 to purchase a third car, and then
trading in this third car at the end of year 4.
Since Sarah wants to minimize her total net cost from now (node 0) to the end of year 4
(node 4), each arc length needs to measure the net cost of that arc’s cycle of purchasing, maintaining, and trading in a car. Therefore,
​Arc length = Purchase price + Operating and maintenance costs − Trade-in value​
For example, consider the arc from node 1 to node 3. This arc corresponds to purchasing a car
at the end of year 1, operating and maintaining it during ownership years 1 and 2, and then
trading it in at the end of ownership year 2. Consequently,
Length of arc from ① to ③ = 12,000 + 2,000 + 3,000 − 6,500
​​
​
​
​​​
​ = 10,500
(in dollars)
The sum of the arc lengths
on any path through this
network gives the total net
cost of the corresponding
plan for trading in cars.
The objective cell now
is TotalCost instead of
TotalDistance.
The arc lengths calculated in this way are shown next to the arcs in Figure 6.14. Adding up
the lengths of the arcs on any path from node 0 to node 4 then gives the total net cost for that
particular plan for trading in cars over the next four years. Therefore, finding the shortest path
from the origin to the destination identifies the plan that will minimize Sarah’s total net cost.
Figure 6.15 shows the corresponding spreadsheet model, formulated in just the same way
as for Figure 6.13 except that distances are now costs. Thus, the objective cell TotalCost (D23)
now gives the total cost that is to be minimized. The changing cells OnRoute (D12:D21) in
the figure display the optimal solution obtained after running Solver. Since values of 1 indicate the path being followed, the shortest path turns out to be
0
2
4
Trade in the first car at the end of year 2.
Trade in the second car at the end of year 4.
The length of this path is 10,500 + 10,500 = 21,000, so Sarah’s total net cost is $21,000, as
given by the objective cell. Recall that this is exactly the amount in Sarah’s car fund provided
by her parents. (These are really nice parents!)
An Example of Minimizing Total Time
The objective is to minimize the total time for the
project.
The Quick Company has learned that a competitor is planning to come out with a new kind
of product with great sales potential. Quick has been working on a similar product that had
been scheduled to come to market in 20 months. However, research is nearly complete and
Quick’s management now wishes to rush the product out to meet the competition.
There are four nonoverlapping phases left to be accomplished, including the remaining
research (the first phase) that currently is being conducted at a normal pace. However, each
phase can instead be conducted at a priority or crash level to expedite completion. These are
the only levels that will be considered for the last three phases, whereas both the normal level
and these two levels will be considered for the first phase. The times required at these levels
are shown in Table 6.3.
Management now has allocated $30 million for these four phases. The cost of each phase
at the levels under consideration is shown in Table 6.4.
Management wishes to determine at which level to conduct each of the four phases to
minimize the total time until the product can be marketed, subject to the budget restriction of
$30 million.
Figure 6.16 shows the network formulation of this problem as a shortest path problem.
Each node indicates the situation at that point in time. Except for the destination, a node is
identified by two numbers:
1. The number of phases completed.
2. The number of millions of dollars left for the remaining phases.
6.4 Shortest Path Problems 225
FIGURE 6.15
A spreadsheet model
that formulates Sarah’s
problem as a shortest path
problem where the objective is to minimize the
total cost instead of the
total distance. The bottom
of the figure shows the
equations entered in the
objective cell TotalCost
(D23) and the other output cells Cost (E12:E21)
and NetFlow (H12:H16).
After applying Solver,
the values of 1 in the
changing cells OnRoute
(D12:D21) identify the
shortest (least expensive) path for scheduling
trade-ins.
A
1
B
C
D
E
F
G
H
I
J
Sarah's Car Purchasing Problem
2
Trade-in
Operating & Value at End Purchase
Maint. Cost
of Year
Price
3
4
5
Year 1
$2,000
$8,500
6
Year 2
$3,000
$6,500
7
Year 3
$4,500
$4,500
8
Year 4
$6,500
$3,000
$12,000
9
10
11
From
To
On Route
Cost
Nodes
Net Flow
12
Year 0
Year 1
0
$5,500
Year 0
1
=
Supply/Demand
1
13
Year 0
Year 2
1
$10,500
Year 1
0
=
0
14
Year 0
Year 3
0
$17,000
Year 2
0
=
0
15
Year 0
Year 4
0
$25,000
Year 3
0
=
0
16
Year 1
Year 2
0
$5,500
Year 4
-1
=
-1
17
Year 1
Year 3
0
$10,500
18
Year 1
Year 4
0
$17,000
19
Year 2
Year 3
0
$5,500
20
Year 2
Year 4
1
$10,500
21
Year 3
Year 4
0
$5,500
Total Cost
$21,000
22
23
Range Name
Cells
Cost
From
NetFlow
Nodes
OnRoute
OpMaint1
OpMaint2
OpMaint3
OpMaint4
PurchasePrice
SupplyDemand
To
TotalCost
TradeIn1
TradeIn2
TradeIn3
TradeIn4
E12:E21
B12:B21
H12:H16
G12:G16
D12:D21
C5
C6
C7
C8
E5
J12:J16
C12:C21
D23
D5
D6
D7
D8
E
Cost
11
12
=PurchasePrice+OpMaint1-TradeIn1
13
=PurchasePrice+OpMaint1+OpMaint2-TradeIn2
14
=PurchasePrice+OpMaint1+OpMaint2+OpMaint3-TradeIn3
15
=PurchasePrice+OpMaint1+OpMaint2+OpMaint3+OpMaint4-TradeIn4
16
=PurchasePrice+OpMaint1-TradeIn1
17
=PurchasePrice+OpMaint1+OpMaint2-TradeIn2
18
=PurchasePrice+OpMaint1+OpMaint2+OpMaint3-TradeIn3
19
=PurchasePrice+OpMaint1-TradeIn1
20
=PurchasePrice+OpMaint1+OpMaint2-TradeIn2
21
=PurchasePrice+OpMaint1-TradeIn1
H
Net Flow
11
Solver Parameters
12
=SUMIF(From,G12,OnRoute)-SUMIF(To,G12,OnRoute)
Set Objective Cell: TotalCost
To: Min
By Changing Variable Cells:
OnRoute
Subject to the Constraints:
NetFlow = SupplyDemand
13
=SUMIF(From,G13,OnRoute)-SUMIF(To,G13,OnRoute)
14
=SUMIF(From,G14,OnRoute)-SUMIF(To,G14,OnRoute)
15
=SUMIF(From,G15,OnRoute)-SUMIF(To,G15,OnRoute)
16
=SUMIF(From,G16,OnRoute)-SUMIF(To,G16,OnRoute)
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
C
23
Total Cost
D
=SUMPRODUCT(OnRoute,Cost)
The origin is now, when 0 phases have been completed and the entire budget of $30 million
is left. Each arc represents the choice of a particular level of effort (identified in parentheses
below the arc) for that phase. [There are no crash arcs emanating from the (2, 12) and (3, 3)
nodes because this level of effort would require exceeding the budget of $30 million for the
four phases.] The time (in months) required to perform the phase with this level of effort
226 Chapter Six Network Optimization Problems
TABLE 6.3
Time Required for
the Phases of Preparing Quick Co.’s New
Product
Level
Remaining
Research
Development
Design of
Manufacturing System
Initiate Production
and Distribution
Normal
Priority
Crash
5 months
4 months
2 months
—
3 months
2 months
—
5 months
3 months
—
2 months
1 month
Level
Remaining
Research
Development
Design of
Manufacturing System
Initiate Production
and Distribution
Normal
Priority
Crash
$3 million
6 million
9 million
—
$6 million
9 million
—
$ 9 million
12 million
—
$3 million
6 million
TABLE 6.4
Cost for the Phases of
Preparing Quick Co.’s
New Product
FIGURE 6.16
Formulation of the Quick
Co. problem as a shortest
path problem. Except for
the dummy destination,
the arc labels indicate,
first, the number of phases
completed and, second,
the amount of money left
(in millions of dollars)
for the remaining phases.
Each arc length gives the
time (in months) to perform that phase.
The sum of the arc lengths
on any path through this
network gives the total
time of the corresponding
plan for preparing the new
product.
5
3
2, 21
(Priority)
y)
t
i
r
o
(Pri
(C 3
ra
1, 27
sh
2
)
(
5
C
)
l
5
r
a
a
s
h
)
m
2, 18
or
(Priority)
3
(Origin)
N
(
ty)
4
iori
(C 3
0, 30
1, 24 (Pr
ra
2
sh
(Priority)
)
(Cra
s
2
5
h
)
(C
2, 15
ra
3
(Priority)
sh
)
ity)
rior
(C 3
1, 21 (P
ra
2
sh
)
(Cra
sh)
5
2, 12
(Priority)
2
(Priority)
(C 1
ra
sh
)
2
3, 9
(Priority)
(C 1
ra
sh
)
2
3, 6
(Priority)
(C 1
ra
sh
)
2
3, 3
(Priority)
3, 12
4, 9
0
4, 6
0
(Destination)
T
0
4, 3
0
4, 0
then is the length of the arc (shown above the arc). Time is chosen as the measure of arc
length because the objective is to minimize the total time for all four phases. Summing the
arc lengths for any particular path through the network gives the total time for the plan corresponding to that path. Therefore, the shortest path through the network identifies the plan
that minimizes total time.
All four phases have been completed as soon as any one of the four nodes with a first label
of 4 has been reached. So why doesn’t the network just end with these four nodes rather than
having an arc coming out of each one? The reason is that a shortest path problem is required
to have only a single destination. Consequently, a dummy destination is added at the righthand side.
When real travel through a network can end at more than one node, an arc with length 0 is
inserted from each of these nodes to a dummy destination so that the network will have just a
single destination.
The objective cell now
is TotalTime instead of
TotalDistance.
Since each of the arcs into the dummy destination has length 0, this addition to the network
does not affect the total length of a path from the origin to its ending point.
Figure 6.17 displays the spreadsheet model for this problem. Once again, the format is the
same as in Figures 6.13 and 6.15, except now the quantity of concern in column F and the
objective cell TotalTime (D32) is time rather than distance or cost. Since Solver has already
been run, the changing cells OnRoute (D4:D30) indicate which arcs lie on the path that minimizes the total time. Thus, the shortest path is
0, 30
1, 21
2, 15
3, 3
4, 0
T
6.4 Shortest Path Problems 227
with a total length of 2 + 3 + 3 + 2 + 0 = 10 months, as given by TotalTime (D32). The
resulting plan for the four phases is shown in Table 6.5. Although this plan does consume the
entire budget of $30 million, it reduces the time until the product can be brought to market
from the originally planned 20 months down to just 10 months.
Given this information, Quick’s management now must decide whether this plan provides
the best trade-off between time and cost. What would be the effect on total time of spending
a few million more dollars? What would be the effect of reducing the spending somewhat
instead? It is easy to provide management with this information as well by quickly solving
FIGURE 6.17
A spreadsheet model that formulates the Quick Co. problem as a shortest path problem where the objective is to minimize the total
time instead of the total distance, so the objective cell is TotalTime (D32). The other output cells are NetFlow (I4:I20). The values of
1 in the changing cells OnRoute (D4:D30) reveal the shortest (quickest) path obtained by Solver.
A
1
B
C
D
E
F
G
H
I
J
K
Quick Co. Product Development Scheduling Problem
2
3
From
To
On Route
Time
Nodes
Net Flow
4
(0, 30)
(1, 27)
0
5
(0, 30)
1
=
1
5
(0, 30)
(1, 24)
0
4
(1, 27)
0
=
0
6
(0, 30)
(1, 21)
1
2
(1, 24)
0
=
0
7
(1, 27)
(2, 21)
0
3
(1, 21)
0
=
0
8
(1, 27)
(2, 18)
0
2
(2, 21)
0
=
0
Supply/Demand
9
(1, 24)
(2, 18)
0
3
(2, 18)
0
=
0
10
(1, 24)
(2, 15)
0
2
(2, 15)
0
=
0
11
(1, 21)
(2, 15)
1
3
(2, 12)
0
=
0
12
(1, 21)
(2, 12)
0
2
(3, 12)
0
=
0
13
(2, 21)
(3, 12)
0
5
(3, 9)
0
=
0
14
(2, 21)
(3, 9)
0
3
(3, 6)
0
=
0
15
(2, 18)
(3, 9)
0
5
(3, 3)
0
=
0
16
(2, 18)
(3, 6)
0
3
(4, 9)
0
=
0
17
(2, 15)
(3, 6)
0
5
(4, 6)
0
=
0
18
(2, 15)
(3, 3)
1
3
(4, 3)
0
=
0
19
(2, 12)
(3, 3)
0
5
(4, 0)
0
=
0
20
(3, 12)
(4, 9)
0
2
(T)
-1
=
-1
21
(3, 12)
(4, 6)
0
1
22
(3, 9)
(4, 6)
0
2
23
(3, 9)
(4, 3)
0
1
24
(3, 6)
(4, 3)
0
2
25
(3, 6)
(4, 0)
0
1
26
(3, 3)
(4, 0)
1
2
27
(4, 9)
(T)
0
0
28
(4, 6)
(T)
0
0
29
(4, 3)
(T)
0
0
(4, 0)
(T)
1
0
30
31
32
Total Time
10
(continued)
228 Chapter Six Network Optimization Problems
FIGURE 6.17
(continued)
Range Name
Cells
From
NetFlow
Nodes
OnRoute
SupplyDemand
Time
To
TotalTime
B4:B30
I4:I20
H4:H20
D4:D30
K4:K20
F4:F30
C4:C30
D32
Solver Parameters
Set Objective Cell: TotalTime
To: Min
By Changing Variable Cells:
OnRoute
Subject to the Constraints:
NetFlow = SupplyDemand
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
I
Net Flow
3
4
=SUMIF(From,H4,OnRoute)-SUMIF(To,H4,OnRoute)
5
=SUMIF(From,H5,OnRoute)-SUMIF(To,H5,OnRoute)
6
=SUMIF(From,H6,OnRoute)-SUMIF(To,H6,OnRoute)
7
=SUMIF(From,H7, OnRoute)-SUMIF(To,H7,OnRoute)
8
=SUMIF(From,H8,OnRoute)-SUMIF(To,H8,OnRoute)
9
=SUMIF(From,H9,OnRoute)-SUMIF(To,H9,OnRoute)
10
=SUMIF(From,H10,OnRoute)-SUMIF(To,H10,OnRoute)
11
=SUMIF(From,H11,OnRoute)-SUMIF(To,H11,OnRoute)
12
=SUMIF(From,H12,OnRoute)-SUMIF(To,H12,OnRoute)
13
=SUMIF(From,H13,OnRoute)-SUMIF(To,H13,OnRoute)
14
=SUMIF(From,H14,OnRoute)-SUMIF(To,H14,OnRoute)
15
=SUMIF(From,H15,OnRoute)-SUMIF(To,H15,OnRoute)
16
=SUMIF(From,H16,OnRoute)-SUMIF(To,H16,OnRoute)
17
=SUMIF(From,H17,OnRoute)-SUMIF(To,H17,OnRoute)
18
=SUMIF(From,H18,OnRoute)-SUMIF(To,H18,OnRoute)
19
=SUMIF(From,H19,OnRoute)-SUMIF(To,H19,OnRoute)
20
=SUMIF(From,H20,OnRoute)-SUMIF(To,H20,OnRoute)
32
C
D
Total Time
=SUMPRODUCT(OnRoute,Time)
some shortest path problems that correspond to budgets different from $30 million. The ultimate decision regarding which plan provides the best time–cost trade-off then is a judgment
decision that only management can make.
TABLE 6.5
The Optimal Solution
Obtained by Solver for
Quick Co.’s Shortest
Path Problem
Phase
Level
Time
Cost
Remaining research
Development
Design of manufacturing system
Initiate production and distribution
Crash
Priority
Crash
Priority
2 months
3 months
3 months
2 months
$ 9 million
6 million
12 million
3 million
10 months
$30 million
Total
Review
Questions
1. What are the origin and the destination in the Littletown Fire Department example?
2. What is the distinction between an arc and a link?
3. What are the supply node and the demand node when a shortest path problem is interpreted
as a minimum-cost flow problem? With what supply and demand?
4. What are three measures of the length of a link (or arc) that lead to three categories of
applications of shortest path problems?
5. What is the objective for Sarah’s shortest path problem?
6. When does a dummy destination need to be added to the formulation of a shortest path
problem?
7. What kind of trade-off does the management of the Quick Co. need to consider in making
its final decision about how to expedite its new product to market?
Chapter 6 Glossary 229
6.5 Summary
Networks of some type arise in a wide variety of contexts. Network representations are very useful for
portraying the relationships and connections between the components of systems. Each component
is represented by a point in the network called a node, and then the connections between components
(nodes) are represented by lines called arcs (for one-way travel) or links (for two-way travel).
Frequently, a flow of some type must be sent through a network, so a decision needs to be made
about the best way to do this. The kinds of network optimization models introduced in this chapter provide a powerful tool for making such decisions.
The model for minimum-cost flow problems plays a central role among these network optimization
models, both because it is so broadly applicable and because it can be readily solved. Solver solves
spreadsheet formulations of reasonable size, and the network simplex method can be used to solve larger
problems, including huge problems with many thousands of nodes and arcs. A minimum-cost flow
problem typically is concerned with optimizing the flow of goods through a network from their points of
origin (the supply nodes) to where they are needed (the demand nodes). The objective is to minimize the
total cost of sending the available supply through the network to satisfy the given demand. One typical
application (among several) is to optimize the operation of a distribution network.
Special types of minimum-cost flow problems include transportation problems and assignment problems (discussed in Chapter 3) as well as two prominent types introduced in this chapter: maximum flow
problems and shortest path problems.
Given the limited capacities of the arcs in the network, the objective of a maximum flow problem is
to maximize the total amount of flow from a particular point of origin (the source) to a particular terminal point (the sink). For example, this might involve maximizing the flow of goods through a company’s
supply network from its vendors to its processing facilities.
A shortest path problem also has a beginning point (the origin) and an ending point (the destination),
but now the objective is to find a path from the origin to the destination that has the minimum total
length. For some applications, length refers to distance, so the objective is to minimize the total distance
traveled. However, some applications instead involve minimizing either the total cost or the total time of
a sequence of activities.
Glossary
arc A channel through which flow may occur
from one node to another, shown as an arrow
between the nodes pointing in the direction in
which flow is allowed. (Section 6.1), 207
assignment problem A special type of
minimum-cost flow problem that previously was
described in Section 3.6. (Section 6.1), 211
capacity of an arc The maximum amount of
flow allowed through the arc. (Section 6.1), 207
conservation of flow Having the amount of
flow out of a node equal the amount of flow into
that node. (Section 6.1), 207
demand node A node where the net amount of
flow generated (outflow minus inflow) is a fixed
negative number, so that flow is absorbed there.
(Section 6.1), 207
destination The node at which travel through
the network is assumed to end for a shortest path
problem. (Section 6.4), 221
dummy destination A fictitious destination
introduced into the formulation of a shortest
path problem with multiple possible termination
points to satisfy the requirement that there be just
a single destination. (Section 6.4), 226
length of a link or arc The number (typically a
distance, a cost, or a time) associated with including the link or arc in the selected path for a shortest path problem. (Section 6.4), 221
link A channel through which flow may
occur in either direction between a pair of
nodes, shown as a line between the nodes.
­(Section 6.4), 220
network simplex method A streamlined
version of the simplex method for solving
­minimum-cost flow problems very efficiently.
(Section 6.1), 210
node A junction point of a network, shown as a
labeled circle. (Section 6.1), 207
origin The node at which travel through the
network is assumed to start for a shortest path
problem. (Section 6.4), 221
sink The node for a maximum flow problem at
which all flow through the network terminates.
(Section 6.3), 216
source The node for a maximum flow problem
at which all flow through the network originates.
(Section 6.3), 216
supply node A node where the net amount of
flow generated (outflow minus inflow) is a fixed
positive number. (Section 6.1), 207
transportation problem A special type of
minimum-cost flow problem that previously
was described in Section 3.5. (Section 6.1),
211
transshipment node A node where the
amount of flow out equals the amount of flow
in. (Section 6.1), 207
transshipment problem A special type of
minimum-cost flow problem where there are no
capacity constraints on the arcs. (Section 6.1), 211
230 Chapter Six Network Optimization Problems
Learning Aids for This Chapter
All learning aids are available at www.mhhe.com/Hillier6e.
Sarah Example
Excel Files:
Quick Example
Distribution Unlimited Example
Excel Add-In:
BMZ Example
Analytic Solver
Expanded BMZ Example
Supplement to This Chapter:
Littletown Fire Department Example
Minimum Spanning-Tree Problems
Solved Problems
The solutions are available at www.mhhe.com/Hillier6e.
6.S1. Distribution at Heart Beats
Heart Beats is a manufacturer of medical equipment. The company’s primary product is a device used to monitor the heart during medical procedures. This device is produced in two factories
and shipped to two warehouses. The product is then shipped on
demand to four third-party wholesalers. All shipping is done by
truck. The product distribution network is shown below. The
annual production capacity at factories 1 and 2 is 400 and 250,
respectively. The annual demand at wholesalers 1, 2, 3, and 4 is
200, 100, 150, and 200, respectively. The cost of shipping one
unit in each shipping lane is shown on the arcs. Because of limited truck capacity, at most 250 units can be shipped from Factory 1 to Warehouse 1 each year. Formulate and solve a network
optimization model in a spreadsheet to determine how to distribute the product at the lowest possible annual cost.
WS
2
F1
$40
[250]
WH
1
$35
$35
WS
2
Exxo 76 is an oil company that operates the pipeline network
shown below, where each pipeline is labeled with its maximum
flow rate in million cubic feet (MMcf) per day. A new oil well
has been constructed near A. They would like to transport oil
from the well near A to their refinery at G. Formulate and solve
a network optimization model to determine the maximum flow
rate from A to G.
F2
$25
[–200]
[–100]
A
20
2
$50
WS
3
[–150]
6.S3. Driving to the Mile-High City
Sarah and Jennifer have just graduated from college at the University of Washington in Seattle and want to go on a road trip.
They have always wanted to see the mile-high city of Denver.
Their road atlas shows the driving time (in hours) between various city pairs, as shown below. Formulate and solve a network
optimization model to find the quickest route from Seattle to
Denver.
Portland
9
7
Butte
10
4
7
7
Boise
6
$65
WS
4
16
15
23
G
4
F
8
C
14
6
12
Seattle
$55
WH
2
B
20
3
[250]
D
15
E
$60
[400]
6.S2. Assessing the Capacity of a Pipeline Network
12
7
14
Salt Lake
5
City
[–200]
Billings
7
Cheyenne
1
Grand 4 Denver
Junction
Chapter 6 Problems 231
Problems
Problems
An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.
6.1.
Read the referenced article that fully describes the management science study summarized in the application vignette
presented in Section 6.1. Briefly describe how the model for a
special type of minimum-cost flow problem was applied in this
study. Then list the various financial and nonfinancial benefits
that resulted from this study.
6.2.*
Consider the transportation problem having the unit
shipping costs shown in the following table.
Destination
1
2
3
Supply
Source
1
2
6
5
7
8
4
6
Demand
30
40
30
40
60
a. Formulate a network model for this problem as a
minimum-cost flow problem by drawing a network
similar to Figure 6.3.
b. Formulate and solve a spreadsheet model for this
problem.
6.3.
Reconsider Problem 6.2. Suppose now that the demand
at each of the destinations has been reduced by 10, so that the
new demands are 20, 30, and 20 at destinations 1, 2, and 3,
respectively. Introduce a dummy destination in order to satisfy
the feasible solutions property and then repeat parts a and b to
determine how much each source should supply to each destination for this new version of the problem,
a. Formulate and solve a spreadsheet model for this
problem without introducing a dummy destination.
b. Now convert this problem into a minimum-flow
cost flow problem by introducing a dummy destination and making the other needed adjustments.
Then draw a network similar to Figure 6.3 for this
minimum-cost flow problem.)
c. Formulate and solve a spreadsheet model for the
minimum-cost flow problem obtained in part b.
Then compare the resulting optimal flow plan
(while ignoring the flows into the dummy destination) with the optimal flow plan obtained in part a
without using a dummy destination.
6.4.
The Makonsel Company is a fully integrated company
that both produces goods and sells them at its retail outlets.
After production, the goods are stored in the company’s two
warehouses until needed by the retail outlets. Trucks are used to
transport the goods from the two plants to the warehouses, and
then from the warehouses to the three retail outlets.
Using units of full truckloads, the first table below shows
each plant’s monthly output, its shipping cost per truckload sent
to each warehouse, and the maximum amount that it can ship per
month to each warehouse.
For each retail outlet (RO), the second table below shows
its monthly demand, its shipping cost per truckload from each
warehouse, and the maximum amount that can be shipped per
month from each warehouse.
Management now wants to determine a distribution plan
(number of truckloads shipped per month from each plant to
each warehouse and from each warehouse to each retail outlet)
that will minimize the total shipping cost.
a. Draw a network that depicts the company’s distribution network. Identify the supply nodes, transshipment nodes, and demand nodes in this network.
b. Formulate a network model for this problem as a
minimum-cost flow problem by inserting all the
necessary data into the network drawn in part a.
(Use the format depicted in Figure 6.3 to display
these data.)
c. Formulate and solve a spreadsheet model for this
problem.
Unit Shipping Cost
Shipping Capacity
To
From
Warehouse 1
Warehouse 2
Warehouse 1
Warehouse 2
Output
Plant 1
Plant 2
$425
510
$560
600
125
175
150
200
200
300
Unit Shipping Cost
Shipping Capacity
To
From
Warehouse 1
Warehouse 2
Demand
RO1
RO2
RO3
RO1
RO2
RO3
$470
390
$505
410
$490
440
100
125
150
150
100
75
150
200
150
150
200
150
232 Chapter Six Network Optimization Problems
6.5.
The Audiofile Company produces boomboxes. However, management has decided to subcontract out the production
of the speakers needed for the boomboxes. Three vendors are
available to supply the speakers. Their price for each shipment of
1,000 speakers is shown below.
Vendor
Price
1
2
3
$22,500
22,700
22,300
Each shipment would go to one of the company’s two warehouses. In addition to the price for each shipment, each vendor
would charge a shipping cost for which it has its own formula
based on the mileage to the warehouse. These formulas and the
mileage data are shown below.
Vendor
Charge per
Shipment
Warehouse 1
Warehouse 2
1
2
3
$300 + 40¢/mile
$200 + 50¢/mile
$500 + 20¢/mile
1,600 miles
500 miles
2,000 miles
400 miles
600 miles
1,000 miles
Whenever one of the company’s two factories needs a shipment of speakers to assemble into the boomboxes, the company
hires a trucker to bring the shipment in from one of the warehouses. The cost per shipment is given next, along with the number of shipments needed per month at each factory.
Unit Shipping Cost
Factory 1
Factory 2
Warehouse 1
$200
$700
Warehouse 2
400
500
10
6
Monthly demand
Each vendor is able to supply as many as 10 shipments per
month. However, because of shipping limitations, each vendor
is only able to send a maximum of six shipments per month to
each warehouse. Similarly, each warehouse is only able to send a
maximum of six shipments per month to each factory.
Management now wants to develop a plan for each month
regarding how many shipments (if any) to order from each vendor, how many of those shipments should go to each warehouse,
and then how many shipments each warehouse should send to
each factory. The objective is to minimize the sum of the purchase costs (including the shipping charge) and the shipping
costs from the warehouses to the factories.
a. Draw a network that depicts the company’s supply
network. Identify the supply nodes, transshipment
nodes, and demand nodes in this network.
b. This problem is only a variant of a minimum-cost flow
problem because the supply from each vendor is a
maximum of 10 rather than a fixed amount of 10. However, it can be converted to a full-fledged minimumcost flow problem by adding a dummy demand node
that receives (at zero cost) all the unused supply
capacity at the vendors. Formulate a network model
for this minimum-cost flow problem by inserting all
the necessary data into the network drawn in part a
supplemented by this dummy demand node. (Use the
format depicted in Figure 6.3 to display these data.)
c. Formulate and solve a spreadsheet model for the
company’s problem.
6.6.*
Consider Figure 6.9 (in Section 6.3), which depicts the
BMZ Co. distribution network from its factories in Stuttgart and
Berlin to the distribution centers in both Los Angeles and Seattle. This figure also gives in brackets the maximum amount that
can be shipped through each shipping lane.
In the weeks following the crisis described in Section 6.2, the
distribution center in Los Angeles has successfully replenished
its inventory. Therefore, Karl Schmidt (the supply chain manager
for the BMZ Co.) has concluded that it will be sufficient hereafter to ship 130 units per month to Los Angeles and 50 units per
month to Seattle. (One unit is a hundred cubic meters of automobile replacement parts.) The Stuttgart factory (node ST in the
figure) will allocate 130 units per month and the Berlin factory
(node BE) will allocate 50 units per month out of their total production to cover these shipments. However, rather than resuming
the past practice of supplying the Los Angeles distribution center
from only the Stuttgart factory and supplying the Seattle distribution center from only the Berlin factory, Karl has decided to allow
either factory to supply either distribution center. He feels that this
additional flexibility is likely to reduce the total shipping cost.
The following table gives the shipping cost per unit through
each of these shipping lanes.
Unit Shipping Cost to Node
To
From
Node
ST
BE
LI
BO
RO
HA
NO
NY
BN
LI
BO
RO
HA
NO
NY
BN
LA
SE
$3,200
—
—
—
—
—
—
—
—
$2,500
—
—
—
—
—
—
—
—
$2,900
$2,400
—
—
—
—
—
—
—
—
$2,000
—
—
—
—
—
—
—
—
—
$6,100
$6,800
—
—
—
—
—
—
—
—
$5,400
$5,900
$6,300
—
—
—
—
—
—
—
—
$5,700
—
—
—
—
—
—
—
—
—
$3,100
$4,200
$3,400
—
—
—
—
—
—
—
$4,000
$3,000
Chapter 6 Problems 233
Karl wants to determine the shipping plan that will minimize
the total shipping cost.
a. Formulate a network model for this problem as a
minimum-cost flow problem by inserting all the
necessary data into the distribution network shown
in Figure 6.9. (Use the format depicted in Figure 6.3
to display these data.)
b. Formulate and solve a spreadsheet model for this
problem.
c. What is the total shipping cost for this optimal
solution?
6.7.
Reconsider Problem 6.6. Suppose now that, for administrative convenience, management has decided that all 130 units
per month needed at the distribution center in Los Angeles must
come from the Stuttgart factory (node ST) and all 50 units per
month needed at the distribution center in Seattle must come
from the Berlin factory (node BE). For each of these distribution
centers, Karl Schmidt wants to determine the shipping plan that
will minimize the total shipping cost.
a. For the distribution center in Los Angeles, formulate a network model for this problem as a
minimum-cost flow problem by inserting all the
necessary data into the distribution network shown
in Figure 6.6. (Use the format depicted in Figure 6.3
to display these data.)
b. Formulate and solve a spreadsheet model for the
problem formulated in part a.
c. For the distribution center in Seattle, draw its distribution network emanating from the Berlin factory at
node BE.
d. Repeat part a for the distribution center in Seattle by
using the network drawn in part c.
e. Formulate and solve a spreadsheet model for the
problem formulated in part d.
f. Add the total shipping costs obtained in parts b and
e. Compare this sum with the total shipping cost
obtained in part c of Problem 6.6 (as given in the
back of the book).
6.8.
Consider the maximum flow problem formulated in
Figures 6.7 and 6.8 for the BMZ case study. Redraw Figure
6.7 and insert the optimal shipping quantities (cells D4:D12 in
­Figure 6.8) in parentheses above the respective arcs. Examine
the capacities of these arcs. Explain why these arc capacities
ensure that the shipping quantities in parentheses must be an
optimal solution because the maximum flow cannot exceed 150.
6.9.
Read the referenced article that fully describes the management science study summarized in the application vignette
presented in Section 6.3. Briefly describe how a generalization
of the model for the maximum flow problem was applied in this
study. Then list the various financial and nonfinancial benefits
that resulted from this study.
6.10.* Formulate and solve a spreadsheet model for the maximum flow problem shown next, where node A is the source,
node F is the sink, and the arc capacities are the numbers in
square brackets shown next to the arcs.
B
[9]
A
D
[7]
[6]
[2]
[3]
F
[4]
[7]
C
[9]
E
[6]
6.11.
The diagram depicts a system of aqueducts that originate at three rivers (nodes R1, R2, and R3) and terminate at a
major city (node T), where the other nodes are junction points in
the system.
A
D
R1
T
B
E
R2
C
R3
F
Using units of thousands of acre feet, the following tables
show the maximum amount of water that can be pumped through
each aqueduct per day.
To
From
R1
R2
R3
A
B
C
75
40
—
65
50
80
—
60
70
To
From
A
B
C
D
E
F
From\To
60
70
—
45
55
70
—
45
90
D
E
F
T
120
190
130
The city water manager wants to determine a flow plan that will
maximize the flow of water to the city.
a. Formulate this problem as a maximum flow problem by identifying a source, a sink, and the transshipment nodes, and then drawing the complete
network that shows the capacity of each arc.
b. Formulate and solve a spreadsheet model for this
problem.
234 Chapter Six Network Optimization Problems
6.12.
The Texago Corporation has four oil fields, four refineries, and four distribution centers in the locations identified
in the next tables. A major strike involving the transportation
industries now has sharply curtailed Texago’s capacity to ship
oil from the four oil fields to the four refineries and to ship petroleum products from the refineries to the distribution centers.
Using units of thousands of barrels of crude oil (and its equivalent in refined products), the following tables show the maximum number of units that can be shipped per day from each oil
field to each refinery and from each refinery to each distribution
center.
Refinery
Oil Field
Texas
California
Alaska
Middle
East
New
Orleans
Charleston
Seattle
St. Louis
11
5
7
8
7
4
3
9
2
8
12
4
8
7
6
15
Distribution Center
Refinery
New Orleans
Charleston
Seattle
St. Louis
Pittsburgh Atlanta
Kansas
City
San
Francisco
5
9
6
4
8
4
12
7
6
11
9
7
9
5
8
7
The Texago management now wants to determine a plan for
how many units to ship from each oil field to each refinery and
from each refinery to each distribution center that will maximize
the total number of units reaching the distribution centers.
a. Draw a rough map that shows the location of Texago’s oil fields, refineries, and distribution centers. Add arrows to show the flow of crude oil and
then petroleum products through this distribution
network.
b. Redraw this distribution network by lining up all the
nodes representing oil fields in one column, all the
nodes representing refineries in a second column,
and all the nodes representing distribution centers in
a third column. Then add arcs to show the possible
flow.
c. Use the distribution network from part b to formulate a network model for Texago’s problem as a
variant of a maximum flow problem.
d. Formulate and solve a spreadsheet model for this
problem.
6.13.
Read the referenced article that fully describes the management science study summarized in the application vignette
presented in Section 6.4. Briefly describe how network optimization models (including for shortest path problems) were
applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study.
6.14.
Reconsider the Littletown Fire Department problem
presented in Section 6.4 and depicted in Figure 6.11. Due to
maintenance work on the one-mile road between nodes A and
B, a detour currently must be taken that extends the trip between
these nodes to four miles.
Formulate and solve a spreadsheet model for this revised
problem to find the new shortest path from the fire station to the
farming community.
6.15.
You need to take a trip by car to another town that you
have never visited before. Therefore, you are studying a map to
determine the shortest route to your destination. Depending on
which route you choose, there are five other towns (call them A,
B, C, D, E) through which you might pass on the way. The map
shows the mileage along each road that directly connects two
towns without any intervening towns. These numbers are summarized in the following table, where a dash indicates that there
is no road directly connecting these two towns without going
through any other towns.
Miles between Adjacent Towns
Town
A
B
C
D
E
Destination
Origin
A
B
C
D
E
40
60
10
50
—
20
—
70
55
—
—
—
40
50
10
—
—
—
—
60
80
a. Formulate a network model for this problem as a
shortest path problem by drawing a network where
nodes represent towns, links represent roads, and
numbers indicate the length of each link in miles.
b. Formulate and solve a spreadsheet model for this
problem.
c. Use part b to identify your shortest route.
d. If each number in the table represented your cost
(in dollars) for driving your car from one town to
the next, would the answer in part c now give your
minimum-cost route?
e. If each number in the table represented your time
(in minutes) for driving your car from one town to
the next, would the answer in part c now give your
minimum-time route?
6.16.* At a small but growing airport, the local airline company is purchasing a new tractor for a tractor-trailer train to
bring luggage to and from the airplanes. A new mechanized luggage system will be installed in three years, so the tractor will
not be needed after that. However, because it will receive heavy
use, so that the running and maintenance costs will increase rapidly as it ages, it may still be more economical to replace the
tractor after one or two years. The next table gives the total net
discounted cost associated with purchasing a tractor (purchase
price minus trade-in allowance, plus running and maintenance
costs) at the end of year i and trading it in at the end of year j
(where year 0 is now).
Case 6-1 Aiding Allies 235
j
3
$8,000
$18,000
10,000
$31,000
21,000
12,000
3.4
SE
3.2
B
E
3.6
LN
3.3
2
4.
Management wishes to determine at what times (if any) the
tractor should be replaced to minimize the total cost for the
tractor(s) over three years.
a. Formulate a network model for this problem as a
shortest path problem.
b. Formulate and solve a spreadsheet model for this
problem.
6.17.
One of Speedy Airlines’s flights is about to take off
from Seattle for a nonstop flight to London. There is some flexibility in choosing the precise route to be taken, depending upon
weather conditions. The following network depicts the possible
routes under consideration, where SE and LN are Seattle and
London, respectively, and the other nodes represent various
intermediate locations.
The winds along each arc greatly affect the flying time (and
so the fuel consumption). Based on current meteorological
4.7
3.6
3.8
2
4.6
1
D
3.4
i
0
1
2
3.5
A
3.5
C
3.4
F
reports, the flying times (in hours) for this particular flight
are shown next to the arcs. Because the fuel consumed is so
expensive, the management of Speedy Airlines has established a policy of choosing the route that minimizes the total
flight time.
a. What plays the role of distances in interpreting this
problem to be a shortest path problem?
b. Formulate and solve a spreadsheet model for this
problem.
Case 6-1
Aiding Allies
Commander Votachev steps into the cold October night and
deeply inhales the smoke from his cigarette, savoring its warmth.
He surveys the destruction surrounding him—shattered windows,
burning buildings, torn roads—and smiles. His two years of work
training revolutionaries east of the Ural Mountains has proven successful; his troops now occupy seven strategically important cities in the Russian Federation: Kazan, Perm, Yekaterinburg, Ufa,
Samara, Saratov, and Orenburg. His siege is not yet over, however.
He looks to the west. Given the political and economic confusion
in the Russian Federation at this time, he knows that his troops will
be able to conquer Saint Petersburg and Moscow shortly. Commander Votachev will then be able to rule with the wisdom and
control exhibited by his communist predecessors Lenin and Stalin.
Across the Pacific Ocean, a meeting of the top security and
foreign policy advisors of the United States is in progress at the
White House. The president has recently been briefed about the
communist revolution masterminded by Commander Votachev
and is determining a plan of action. The president reflects upon
a similar October long ago in 1917, and he fears the possibility
of a new age of radical Communist rule accompanied by chaos,
bloodshed, escalating tensions, and possibly nuclear war. He
therefore decides that the United States needs to respond and
to respond quickly. Moscow has requested assistance from the
United States military, and the president plans to send troops and
supplies immediately.
The president turns to General Lankletter and asks him to
describe the preparations being taken in the United States to send
the necessary troops and supplies to the Russian Federation.
General Lankletter informs the president that along with
troops, weapons, ammunition, fuel, and supplies, aircraft,
ships, and vehicles are being assembled at two port cities with
airfields: Boston and Jacksonville. The aircraft and ships will
transfer all troops and cargo across the Atlantic Ocean to the
Eurasian continent. The general hands the president a list of the
types of aircraft, ships, and vehicles being assembled along with
a description of each type. The list is shown next.
Transportation Type
Name
Capacity
Speed
Aircraft
Ship
Vehicle
C-17 Globemaster
Transport
Palletized Load
System Truck
150 tons
240 tons
16,000 kilograms
400 miles per hour
35 miles per hour
60 miles per hour
236 Chapter Six Network Optimization Problems
All aircraft, ships, and vehicles are able to carry both troops
and cargo. Once an aircraft or ship arrives in Europe, it stays
there to support the armed forces.
The president then turns to Tabitha Neal, who has been negotiating with the NATO countries for the last several hours to use
their ports and airfields as stops to refuel and resupply before
heading to the Russian Federation. She informs the president
that the following ports and airfields in the NATO countries will
be made available to the U.S. military.
Ports
Airfields
Napoli
Hamburg
Rotterdam
London
Berlin
Istanbul
The president stands and walks to the map of the world projected on a large screen in the middle of the room. He maps the
progress of troops and cargo from the United States to three
strategic cities in the Russian Federation that have not yet been
seized by Commander Votachev. The three cities are Saint
Petersburg, Moscow, and Rostov. He explains that the troops
and cargo will be used both to defend the Russian cities and
to launch a counter attack against Votachev to recapture the
cities he currently occupies. (The map is shown at the end of
the case.)
The president also explains that all Globemasters and transports leave Boston or Jacksonville. All transports that have traveled across the Atlantic must dock at one of the NATO ports
to unload. Palletized load system trucks brought over in the
transports will then carry all troops and materials unloaded from
the ships at the NATO ports to the three strategic Russian cities
not yet seized by Votachev. All Globemasters that have traveled
across the Atlantic must land at one of the NATO airfields for
refueling. The planes will then carry all troops and cargo from
the NATO airfields to the three Russian cities.
a. Draw a network showing the different routes troops and
supplies may take to reach the Russian Federation from the
United States.
b. Moscow and Washington do not know when Commander
Votachev will launch his next attack. Leaders from the two
countries therefore have agreed that troops should reach each
of the three strategic Russian cities as quickly as possible. The
president has determined that the situation is so dire that cost
is no object—as many Globemasters, transports, and trucks as
are necessary will be used to transfer troops and cargo from
the United States to Saint Petersburg, Moscow, and Rostov.
Therefore, no limitations exist on the number of troops and
amount of cargo that can be transferred between any cities.
The president has been given the information in the next
table about the length of the available routes between cities.
Given the distance and the speed of the transportation
used between each pair of cities, how can the president most
quickly move troops from the United States to each of the
three strategic Russian cities? Highlight the path(s) on the
network. How long will it take troops and supplies to reach
Saint Petersburg? Moscow? Rostov?
From
To
(Kilometers)
Boston
Boston
Boston
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Saint Petersburg
Saint Petersburg
Saint Petersburg
Saint Petersburg
Saint Petersburg
Saint Petersburg
Moscow
Moscow
Moscow
Moscow
Moscow
Moscow
Rostov
Rostov
Rostov
Rostov
Rostov
Rostov
7,250 km
8,250
8,300
6,200
6,900
7,950
9,200
9,800
10,100
7,900
8,900
9,400
1,280
1,880
2,040
1,980
2,200
2,970
1,600
2,120
1,700
2,300
2,450
2,890
1,730
2,470
990
2,860
2,760
2,800
c. The president encounters only one problem with his first
plan: He has to sell the military deployment to Congress.
Under the War Powers Act, the president is required to consult with Congress before introducing troops into hostilities
or situations where hostilities will occur. If Congress does
not give authorization to the president for such use of troops,
the president must withdraw troops after 60 days. Congress
also has the power to decrease the 60-day time period by
passing a concurrent resolution.
The president knows that Congress will not authorize significant spending for another country’s war, especially when
voters have paid so much attention to decreasing the national
debt. He therefore decides that he needs to find a way to get
the needed troops and supplies to Saint Petersburg, Moscow,
and Rostov at the minimum cost.
Each Russian city has contacted Washington to communicate the number of troops and supplies the city needs at
a minimum for reinforcement. After analyzing the requests,
General Lankletter has converted the requests from numbers
of troops, gallons of gasoline, and so on, to tons of cargo for
easier planning. The requirements are listed next.
City
Saint Petersburg
Moscow
Rostov
Requirements
320,000 tons
440,000 tons
240,000 tons
Case 6-1 Aiding Allies 237
Both in Boston and Jacksonville, there are 500,000 tons of
the necessary cargo available. When the United States decides
to send a plane, ship, or truck between two cities, several
costs occur: fuel costs, labor costs, maintenance costs, and
appropriate port or airfield taxes and tariffs. These costs are
listed next.
From
To
Cost
Boston
Boston
Boston
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Jacksonville
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Berlin
Hamburg
Istanbul
London
Rotterdam
Napoli
Saint Petersburg
Saint Petersburg
Saint Petersburg
Saint Petersburg
Saint Petersburg
Saint Petersburg
Moscow
Moscow
Moscow
Moscow
Moscow
Moscow
Rostov
Rostov
Rostov
Rostov
Rostov
Rostov
$50,000 per Globemaster
$30,000 per transport
$55,000 per Globemaster
$45,000 per Globemaster
$30,000 per transport
$32,000 per transport
$57,000 per Globemaster
$48,000 per transport
$61,000 per Globemaster
$49,000 per Globemaster
$44,000 per transport
$56,000 per transport
$24,000 per Globemaster
$3,000 per truck
$28,000 per Globemaster
$22,000 per Globemaster
$3,000 per truck
$5,000 per truck
$22,000 per Globemaster
$4,000 per truck
$25,000 per Globemaster
$19,000 per Globemaster
$5,000 per truck
$5,000 per truck
$23,000 per Globemaster
$7,000 per truck
$2,000 per Globemaster
$4,000 per Globemaster
$8,000 per truck
$9,000 per truck
The president faces a number of restrictions when trying
to satisfy the requirements. Early winter weather in northern
Russia has brought a deep freeze with much snow. Therefore,
General Lankletter is opposed to sending truck convoys in the
area. He convinces the president to supply Saint Petersburg only
through the air. Moreover, the truck routes into Rostov are quite
limited, so that from each port, at most 2,500 trucks can be sent
to Rostov. The Ukrainian government is very sensitive about
American airplanes flying through its air space. It restricts the
U.S. military to at most 200 flights from Berlin to Rostov and to
at most 200 flights from London to Rostov. (The U.S. military
does not want to fly around the Ukraine and is thus restricted by
the Ukrainian limitations.)
How does the president satisfy each Russian city’s military
requirements at minimum cost? Highlight the path to be used
between the United States and the Russian Federation on the
network.
d. Once the president releases the number of planes, ships, and
trucks that will travel between the United States and the
Russian Federation, Tabitha Neal contacts each of the
American cities and NATO countries to indicate the number
of planes to expect at the airfields, the number of ships to
expect at the docks, and the number of trucks to expect traveling across the roads. Unfortunately, Tabitha learns that
several additional restrictions exist that cannot be immediately eliminated. Because of airfield congestion and unalterable flight schedules, only a limited number of planes may
be sent between any two cities. These plane limitations are
given below.
From
To
Boston
Boston
Boston
Jacksonville
Jacksonville
Jacksonville
Berlin
Istanbul
London
Berlin
Istanbul
London
Berlin
Istanbul
London
Berlin
Istanbul
London
Berlin
Istanbul
London
Saint Petersburg
Saint Petersburg
Saint Petersburg
Moscow
Moscow
Moscow
Rostov
Rostov
Rostov
Maximum
Number of Airplanes
300
500
500
500
700
600
500
0
1,000
300
100
200
0
900
100
In addition, because some countries fear that citizens will
become alarmed if too many military trucks travel the public
highways, they object to a large number of trucks traveling
through their countries. These objections mean that a limited
number of trucks are able to travel between certain ports and
Russian cities. These limitations are listed below.
From
To
Rotterdam
Rotterdam
Hamburg
Hamburg
Napoli
Napoli
Moscow
Rostov
Moscow
Rostov
Moscow
Rostov
Maximum
Number of Trucks
600
750
700
500
1,500
1,400
Tabitha learns that all shipping lanes have no capacity limits
due to the American control of the Atlantic Ocean.
The president realizes that due to all the restrictions, he will
not be able to satisfy all the reinforcement requirements of the
three Russian cities. He decides to disregard the cost issue and
instead to maximize the total amount of cargo he can get to
the Russian cities. How does the president maximize the total
amount of cargo that reaches the Russian Federation? Highlight
the path(s) used between the United States and the Russian Federation on the network.
238 Chapter Six Network Optimization Problems
St.
Petersburg
Boston
London
Jacksonville
Hamburg Berlin
Rotterdam
Perm
Kazan Yekaterinburg
Moscow
Ufa
Samara
Orenburg
Saratov
Rostov
Napoli
Istanbul
Case 6-2
Money in Motion
Jake Nguyen runs a nervous hand through his once finely
combed hair. He loosens his once perfectly knotted silk tie. And
he rubs his sweaty hands across his once immaculately pressed
trousers. Today has certainly not been a good day.
Over the past few months, Jake had heard whispers circulating from Wall Street—whispers from the lips of investment
bankers and stockbrokers famous for their outspokenness. They
had whispered about a coming Japanese economic collapse—
whispered because they had believed that publicly vocalizing
their fears would hasten the collapse.
And, today, their very fears have come true. Jake and his
colleagues gather around a small television dedicated exclusively to the Bloomberg channel. Jake stares in disbelief as he
listens to the horrors taking place in the Japanese market. And
the Japanese market is taking the financial markets in all other
East Asian countries with it on its tailspin. He goes numb. As
manager of Asian foreign investment for Grant Hill Associates,
a small West Coast investment boutique specializing in currency trading, Jake bears personal responsibility for any negative
impacts of the collapse. And Grant Hill Associates will experience negative impacts.
Jake had not heeded the whispered warnings of a Japanese
collapse. Instead, he had greatly increased the stake Grant Hill
Associates held in the Japanese market. Because the Japanese
market had performed better than expected over the past year,
Jake had increased investments in Japan from $2.5 million to
$15 million only one month ago. At that time, one dollar was
worth 80 yen.
No longer. Jake realizes that today’s devaluation of the yen
means that one dollar is worth 125 yen. He will be able to liquidate these investments without any loss in yen, but now the dollar loss when converting back into U.S. currency would be huge.
He takes a deep breath, closes his eyes, and mentally prepares
himself for serious damage control.
Jake’s meditation is interrupted by a booming voice calling
for him from a large, corner office. Grant Hill, the president of
Grant Hill Associates, yells, “Nguyen, get the hell in here!”
Jake jumps and looks reluctantly toward the corner office
hiding the furious Grant Hill. He smooths his hair, tightens his
tie, and walks briskly into the office.
Grant Hill meets Jake’s eyes upon his entrance and continues
yelling, “I don’t want one word out of you, Nguyen! No excuses;
just fix this debacle! Get all of our money out of Japan! My gut
tells me this is only the beginning! Get the money into safe U.S.
bonds! NOW! And don’t forget to get our cash positions out of
Indonesia and Malaysia ASAP with it!”
Jake has enough common sense to say nothing. He nods his
head, turns on his heels, and practically runs out of the office.
Safely back at his desk, Jake begins formulating a plan to
move the investments out of Japan, Indonesia, and Malaysia.
His experiences investing in foreign markets have taught him
that when playing with millions of dollars, how he gets money
out of a foreign market is almost as important as when he gets
money out of the market. The banking partners of Grant Hill
Associates charge different transaction fees for converting one
currency into another one and wiring large sums of money
around the globe.
And now, to make matters worse, the governments in East
Asia have imposed very tight limits on the amount of money an
individual or a company can exchange from the domestic currency into a particular foreign currency and withdraw it from
the country. The goal of this dramatic measure is to reduce the
outflow of foreign investments out of those countries to prevent
a complete collapse of the economies in the region. Because of
Grant Hill Associates’ cash holdings of 10.5 billion Indonesian
rupiahs and 28 million Malaysian ringgits, along with the holdings in yen, it is not clear how these holdings should be converted back into dollars.
Case 6-2 Money in Motion 239
Jake wants to find the most cost-effective method to convert
these holdings into dollars. On his company’s website, he always
can find on-the-minute exchange rates for most currencies in the
world (see Table 1).
The table states that, for example, 1 Japanese yen equals
0.008 U.S. dollars. By making a few phone calls, he discovers
the transaction costs (expressed as a percentage of the currency
being converted) his company must pay for large currency transactions during these critical times (see Table 2).
Jake notes that exchanging one currency for another one
results in the same transaction cost as a reverse conversion.
Finally, Jake finds out the maximum amounts of domestic currencies (expressed as the current equivalent of thousands of dollars) his company is allowed to convert into other currencies in
Japan, Indonesia, and Malaysia (see Table 3).
a. Formulate Jake’s problem as a minimum-cost flow problem,
and draw the network for his problem. Identify the supply
and demand nodes for the network.
b. Which currency transactions must Jake perform to convert
the investments from yens, rupiahs, and ringgits into U.S.
dollars to ensure that Grant Hill Associates has the maximum
dollar amount after all transactions have occurred? How
much money does Jake have to invest in U.S. bonds?
c. The World Trade Organization forbids transaction limits
because they promote protectionism. If no transaction limits exist, what method should Jake use to convert the Asian
holdings from the respective currencies into dollars?
d. In response to the World Trade Organization’s mandate
forbidding transaction limits, the Indonesian government
introduces a new tax to protect its currency that leads to a
500 percent increase in transaction costs for transactions of
rupiahs. Given these new transaction costs but no transaction limits, what currency transactions should Jake perform
to convert the Asian holdings from the respective currencies
into dollars?
e. Jake realizes that his analysis is incomplete because he has
not included all aspects that might influence his planned
currency exchanges. Describe other factors that Jake should
examine before he makes his final decision.
TABLE 1
Currency Exchange Rates
To
From
Yen
Rupiah
Ringgit
U.S. Dollar
Canadian Dollar
Euro
Pound
Peso
1
50
1
0.04
0.0008
1
0.008
0.00016
0.2
1
0.01
0.0002
0.25
1.25
1
0.0064
0.000128
0.16
0.8
0.64
1
0.0048
0.000096
0.12
0.6
0.48
0.75
1
0.0768
0.001536
1.92
9.6
7.68
12
16
1
Japanese yen
Indonesian rupiah
Malaysian ringgit
U.S. dollar
Canadian dollar
European euro
English pound
Mexican peso
TABLE 2
Transaction Cost (Percent)
To
From
Yen
Rupiah
Ringgit
U.S. Dollar
Canadian Dollar
Euro
Pound
Peso
—
0.5
—
0.5
0.7
—
0.4
0.5
0.7
—
0.4
0.3
0.7
0.05
—
0.4
0.3
0.4
0.1
0.2
—
0.25
0.75
0.45
0.1
0.1
0.05
—
0.5
0.75
0.5
0.1
0.1
0.5
0.5
—
Yen
Rupiah
Ringgit
U.S. dollar
Canadian dollar
Euro
Pound
Peso
TABLE 3
Transaction Limits in Equivalent of 1,000 Dollars
To
From
Yen
Rupiah
Ringgit
U.S. Dollar
Canadian Dollar
Euro
Pound
Peso
Yen
Rupiah
Ringgit
—
5,000
3,000
5,000
—
4,500
5,000
2,000
—
2,000
200
1,500
2,000
200
1,500
2,000
1,000
2,500
2,000
500
1,000
4,000
200
1,000
240 Chapter Six Network Optimization Problems
Case 6-3
Airline Scheduling
Rachel Cook is very concerned. Until recently, she has always
had the golden touch, having successfully launched two startup companies that made her a very wealthy woman. However,
the timing could not have been worse for her latest start-up—
a regional airline called Northwest Commuter that operates
on the west coast of the United States. All had been well at
the beginning. Four airplanes had been leased and the company had become fairly well established as a no-frills airline
providing low-cost commuter flights between the west coast
cities of Seattle, Portland, and San Francisco. Achieving fast
turnaround times between flights had given Northwest Commuter an important competitive advantage. Then the cost of
jet fuel began spiraling upward and the company began going
heavily into the red (like so many other airlines at the time).
Although some of the flights were still profitable, others
were losing a lot of money. Fortunately, jet fuel costs now
are starting to come down, but it has become clear to Rachel
that she needs to find new ways for Northwest Commuter to
become a more efficient airline. In particular, she wants to
start by dropping unprofitable flights and then identifying the
most profitable combination of flights (including some new
ones) for the coming year that could feasibly be flown by the
four airplanes.
A little over a decade ago, Rachel had been an honor graduate of a leading MBA program. She had enjoyed the management science course she took then and she has decided to apply
spreadsheet modeling to analyze her problem.
The leasing cost for each airplane is $30,000 per day. At the
end of the day, an airplane might remain in the city where it
landed on its last flight. Another option is to fly empty overnight
to another city to be ready to start a flight from there the next
morning. The cost of this latter option is $5,000.
Flight Number
From
To
Depart
Arrive
1257
2576
8312
1109
3752
2498
8787
8423
7922
5623
2448
1842
3487
4361
4299
1288
3335
9348
7400
7328
6386
6923
Seattle
Seattle
Seattle
Seattle
Seattle
Seattle
Seattle
Seattle
Portland
Portland
Portland
Portland
Portland
Portland
Portland
San Francisco
San Francisco
San Francisco
San Francisco
San Francisco
San Francisco
San Francisco
San Francisco
Portland
San Francisco
San Francisco
San Francisco
Portland
San Francisco
Portland
Seattle
San Francisco
San Francisco
Seattle
Seattle
San Francisco
Seattle
Seattle
Portland
Seattle
Seattle
Portland
Portland
Seattle
8:00 AM
9:30 AM
9:30 AM
12:00 PM
2:30 PM
3:00 PM
5:00 PM
6:30 PM
9:00 AM
9:30 AM
11:00 AM
12:00 PM
2:00 PM
4:00 PM
6:00 PM
8:00 AM
8:30 AM
10:30 AM
12:00 PM
12:00 PM
4:00 PM
5:00 PM
10:00 AM
10:30 AM
11:30 AM
2:00 PM
4:30 PM
4:00 PM
7:00 PM
7:30 PM
10:00 AM
11:00 AM
12:30 PM
1:00 PM
3:00 PM
5:30 PM
7:00 PM
10:00 AM
10:00 AM
12:30 PM
2:00 PM
1:30 PM
5:30 PM
7:00 PM
The accompanying table shows the 22 possible flights that
are being considered for the coming year. The last column gives
the estimated net revenue (in thousands of dollars) for each
flight, given the average number of passengers anticipated for
that flight.
a. To simplify the analysis, assume for now that there is virtually no turnaround time between flights so the next flight
Expected
Revenue ($000)
37
20
25
27
23
18
29
27
20
23
19
21
22
29
27
32
26
24
27
24
28
32
can begin as soon as the current flight ends. (If an immediate
next flight is not available, the airplane would wait until the
next scheduled flight from that city.) Develop a network that
displays some of the feasible routings of the flights. (Hint:
Include separate nodes for each half hour between 8:00 AM
and 7:30 PM in each city.) Then develop and apply the corresponding spreadsheet model that finds the feasible combination of flights that maximizes the total profit.
Case 6-4 Broadcasting the Olympic Games 241
b. Rachel is considering leasing additional airplanes to achieve
economies of scale. The leasing cost of each one again would
be $30,000 per day. Perform what-if analysis to determine
whether it would be worthwhile to have 5, 6, or 7 airplanes
instead of 4.
c. Now repeat part a under the more realistic assumption that
there is a minimum turnaround time of 30 minutes on the
ground for unloading and loading passengers between the
arrival of a flight and the departure of the next flight by
the same airplane. (Most airlines use a considerably longer
turnaround time.) Does this change the number of flights that
can be flown?
d. Rachel now is considering having each of the four airplanes
carry freight instead of flying empty if it flies overnight to
another city. Instead of a cost of $5,000, this would result in
net revenue of $5,000. Adapt the spreadsheet model used in
part c to find the feasible combination of flights that maximizes the total profit. Does this change the number of airplanes that fly overnight to another city?
Case 6-4
Broadcasting the Olympic Games
The management of the WBC television network has been
celebrating for days. What a coup! After several unsuccessful
attempts in recent decades, they finally have hit the big jackpot.
They have won the bidding war to gain the rights to broadcast
the next Summer Olympic Games!
The price was enormous. However, the advertising income
also will be huge. Even if the network loses some money in the
process, the gain in prestige should make it all worthwhile. After
all, the entire world follows these games closely every four years.
Now the entire world receiving the feed of the broadcast from
the WBC network will learn what a preeminent network it is.
However, reality also is setting in for WBC management.
Telecasting the entire Olympic Games will be an enormously
complex task. Many different sporting events will be occurring
simultaneously in far-flung venues. An unprecedented amount
of live television and live-on-the-Internet coverage of the various sporting events needs to be planned.
Due to the high amount of bandwidth that will be required
to transmit the coverage of the games back to its home studios,
WBC needs to upgrade its computer network. It operates a private computer network as shown in the network diagram below.
The games will be held near node A. WBC’s home studios are
located at node G. At peak times, coverage of the games will
require 35 GB/s (35 gigabytes of data per second) to be sent
through the network from node A to node G. The capacity of
each link in the network is shown in the diagram below (in
GB/s). WBC can divide the transmission and route it through
multiple paths of the network from A to G, so long as the total
bandwidth required on each link does not exceed the capacity
of that link.
7
B
13
5
A
E
9
D
C
a. By utilizing the entire computer network, what is the maximum bandwidth available (in GB/s) for transmission from
the general site of the Olympic Games (node A) to the home
studios (node G)? Set up and solve a linear programming
spreadsheet model.
b. WBC would like to expand the capacity of the network so it
can handle the peak requirement of 35 GB/s from the Olympics site (A) to the home studios (G). WBC can increase the
capacity of each link of the computer network by installing
9
4
6
G
3
10
6
F
12
8
additional fiber optic cables. The next table shows the existing capacity of each network segment (in GB/s), the maximum additional capacity that can be added (in GB/s), and the
cost to increase the capacity (in millions of dollars per unit
GB/s added). Make a copy of the spreadsheet model used to
solve part a and make any revisions necessary to solve this
new problem of handling the peak requirement of 35 GB/s.
Note: This case will be continued in the next chapter ( Case 7-4 ),
so we suggest that you save your spreadsheet model from part b.
242 Chapter Six Network Optimization Problems
Network Segment
From
To
Existing Capacity
(GB/s)
Maximum Additional
Capacity (GB/s)
Cost per GB/s of Additional
Capacity ($million)
A
A
A
B
B
B
C
D
D
E
E
F
B
C
D
D
E
F
D
E
G
F
G
G
13
6
10
9
5
7
8
3
12
4
6
9
6
4
3
4
5
3
5
2
5
2
4
5
2.8
2.5
2.8
2.5
3.1
1.6
3.9
2.8
1.6
4.6
2.9
1.8
Additional Cases
Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases.
ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book.
Chapter Seven
Using Binary Integer
Programming to Deal
with Yes-or-No Decisions
Learning Objectives
After completing this chapter, you should be able to
1. Describe how binary decision variables are used to represent yes-or-no decisions.
2. Use binary decision variables to formulate constraints for mutually exclusive alternatives
and contingent decisions.
3. Formulate a binary integer programming model for the selection of projects.
4. Formulate a binary integer programming model for the selection of sites for facilities.
5. Formulate a binary integer programming model for crew scheduling in the travel industry.
6. Formulate other basic binary integer programming models from a description of the
problems.
7. Use mixed binary integer programming to deal with setup costs for initiating the production
of a product.
The preceding chapters have considered various kinds of problems where decisions need to be
made about how much to do of various activities. Thus, the decision variables in the resulting
model represent the level of the corresponding activities.
We turn now to a common type of problem where, instead of how-much decisions, the
decisions to be made are yes-or-no decisions. A yes-or-no decision arises when a particular
option is being considered and the only possible choices are yes, go ahead with this option, or
no, decline this option.
The natural choice of a decision variable for a yes-or-no decision is a binary variable.
Binary variables are variables whose only possible values are 0 and 1. Thus, when representing a yes-or-no decision, a binary decision variable is assigned a value of 1 for choosing yes and a value of 0 for choosing no.
Models that fit linear programming except that they use binary decision variables are called
binary integer programming (BIP) models. (We hereafter will use the BIP abbreviation.)
A pure BIP model is one where all the variables are binary variables, whereas a mixed BIP
model is one where only some of the variables are binary variables.
A BIP model can be considered to be a special type of integer programming model. A general integer programming model is simply a linear programming model except for also having
constraints that some or all of the decision variables must have integer values (0, 1, 2, . . .). A
BIP model further restricts these integer values to be only 0 or 1.
However, BIP problems are quite different from general integer programming problems
because of the difference in the nature of the decisions involved. Like linear programming
problems, general integer programming problems involve how-much decisions, but where
these decisions make sense only if they have integer values. For example, the TBA Airlines
problem presented in Section 3.2 is a general integer programming problem because it is
243
244 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
essentially a linear programming problem except that its how-much decisions (how many
small airplanes and how many large airplanes to purchase) only make sense if they have integer values. By contrast, pure BIP problems involve yes-or-no decisions instead of how-much
decisions. Mixed BIP problems involve both types of decisions.
The first four sections of the chapter present a variety of examples of pure BIP problems.
Section 7.5 provides an example of a mixed BIP problem. For clarity, all of these examples
have been kept quite small, so sometimes they could be solved by simply enumerating and
evaluating the options individually without even needing BIP. However, they are illustrative
of the sometimes vastly larger problems that often arise in practice and require the use of BIP.
The preceding chapters already have focused on problems involving how-much decisions
and how such techniques as linear programming or integer programming can be used to a­ nalyze
these problems. Therefore, this chapter will be devoted instead to problems involving yes-or-no
decisions and how BIP models can be used to analyze this special category of problems.
BIP problems arise with considerable frequency in a wide variety of applications. To illustrate this, we begin with a case study and then present some more examples in the subsequent
sections. One of the supplements to this chapter at www.mhhe.com/Hillier6e also provides
additional formulation examples for BIP problems.
You will see throughout this chapter that BIP problems can be formulated on a spreadsheet
just as readily as linear programming problems. The Solver also can solve BIP problems of
modest size. You normally will have no problem solving the small BIP problems found in
this book, but Solver may fail on somewhat larger problems. To provide some perspective on
this issue, we include another supplement at www.mhhe.com/Hillier6e that is entitled Some
Perspectives on Solving Binary Integer Programming Problems. The algorithms available for
solving BIP problems (including the one used by Solver) are not nearly as efficient as those
for linear programming, so this supplement discusses some of the difficulties and pitfalls
involved in solving large BIP problems. One option with any large problem that fits linear
programming except that it has decision variables that are restricted to integer values (but
not necessarily just 0 and 1) is to ignore the integer constraints and then to round the solution
obtained to integer values. This is a reasonable option in some cases but not in others. The
supplement emphasizes that this is a particularly dangerous shortcut with BIP problems.
7.1
A CASE STUDY: THE CALIFORNIA MANUFACTURING CO. PROBLEM
The top management of the California Manufacturing Company wants to develop a plan for
the expansion of the company. Therefore, a management science study will be conducted
to help guide the decisions that need to be made. The president of the company, Armando
Ortega, is about to meet with the company’s top management scientist, Steve Chan, to discuss
the study that management wants done. Let’s eavesdrop on this meeting.
Armando Ortega (president): OK, Steve, here is the situation. With our growing business,
we are strongly considering building a new factory. Maybe even two. The factory needs to be
close to a large, skilled labor force, so we are looking at Los Angeles and San Francisco as
the potential sites. We also are considering building one new warehouse. Not more than one.
This warehouse would make sense in saving shipping costs only if it is in the same city as a
new factory. Either Los Angeles or San Francisco. If we decide not to build a new factory at
all, we definitely don’t want the warehouse either. Is this clear, so far?
Steve Chan (management scientist): Yes, Armando, I understand, What are your criteria
for making these decisions?
Armando Ortega: Well, all the other members of top management have joined me in addressing
this issue. We have concluded that these two potential sites are very comparable on nonfinancial
grounds. Therefore, we feel that these decisions should be based mainly on financial considerations. We have $10 million of capital available for this expansion and we want it to go as far as
possible in improving our bottom line. Which feasible combination of investments in factories
and warehouses in which locations will be most profitable for the company in the long run?
In your language, we want to maximize the total net present value of these investments.
An Application Vignette
With headquarters in Houston, Texas, Waste Management, Inc.
(a Fortune 100 company), is the leading provider of comprehensive waste-management services in North America. Its network
of operations includes hundreds of active landfill disposal sites,
recycling plants, transfer stations, and collection operations
(depots) to provide services to over 25 million residential commercial customers throughout the United States and Canada.
The company’s collection-and-transfer vehicles need to follow tens of thousands of daily routes. With an annual operating cost of approximately $120,000 per vehicle, management
wanted to have a comprehensive route-management system
that would make every route as profitable and efficient as possible. Therefore, a management science team that included a
number of consultants was formed to attack this problem.
The heart of the route-management system developed by this
team is a huge mixed BIP model that optimizes the routes assigned
to the respective collection-and-transfer vehicles. Although the
What is the most profitable combination of
investments?
objective function takes several factors into account, the primary
goal is the minimization of total travel time. The main decision variables are binary variables that equal 1 if the route assigned to a
particular vehicle includes a particular possible leg and that equal
0 otherwise. A geographical information system (GIS) provides the
data about the distance and time required to go between any two
points. All of this is imbedded within a Web-based Java application
that is integrated with the company’s other systems.
It is estimated that the implementation of this comprehensive route-management system increased the company’s cash
flow by $648 million over a five-year period, largely because of
savings of $498 million in operational expenses over this same
period. It also is providing better customer service.
Source: S. Sahoo, S. Kim, B.-I. Kim, B. Krass, and A. Popov, Jr.,
“Routing Optimization for Waste Management,” Interfaces 35, no. 1
(January–February 2005), pp. 24–36. (A link to this article is provided
at www.mhhe.com/Hillier6e.)
Steve Chan: That’s very clear. It sounds like a classical management science problem.
Armando Ortega: That’s why I called you in, Steve. I would like you to conduct a quick
management science study to determine the most profitable combination of investments.
I also would like you to take a look at the amount of capital being made available and
its effect on how much profit we can get from these investments. The decision to make
$10 million available is only a tentative one. That amount is stretching us, because we now
are investigating some other interesting project proposals that would require quite a bit of
capital, so we would prefer to use less than $10 million on these particular investments if
the last few million don’t buy us much. On the other hand, this expansion into either Los
Angeles or San Francisco, or maybe both of these key cities, is our number one priority. It
will have a real positive impact on the future of this company. So we are willing to go out
and raise some more capital if it would give us a lot of bang for the buck. Therefore, we
also would like you to do some what-if analysis to tell us what the effect would be if we
were to change the amount of capital being made available to anything between $5 million
and $15 million.
Steve Chan: Sure, Armando, we do that kind of what-if analysis all the time. We refer to
it as sensitivity analysis because it involves checking how sensitive the outcome is to the
amount of capital being made available.
Armando Ortega: Good. Now, Steve, I need your input within the next couple weeks.
Can you do it?
Steve Chan: Well, Armando, as usual, the one question is whether we can gather all the
necessary data that quickly. We’ll need to get good estimates of the net present value of each
of the possible investments. I’ll need a lot of help in digging out that information.
Armando Ortega: I thought you would say that. I already have my staff working hard on
developing those estimates. I can get you together with them this afternoon.
Steve Chan: Great. I’ll get right on it.
Background
The California Manufacturing Company is a diversified company with several factories
and warehouses throughout California, but none yet in Los Angeles or San Francisco. Because
the company is enjoying increasing sales and earnings, management feels that the time may
be ripe to expand into one or both of those prime locations. A basic issue is whether to build
a new factory in either Los Angeles or San Francisco, or perhaps even in both cities. Management also is considering building at most one new warehouse, but will restrict the choice of
location to a city where a new factory is being built.
245
246 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
TABLE 7.1
Data for the California
Manufacturing Co.
Problem
Decision
Number
Yes-or-No
Question
Decision
Variable
Net Present
Value (Millions)
1
2
3
4
Build a factory in Los Angeles?
Build a factory in San Francisco?
Build a warehouse in Los Angeles?
Build a warehouse in San Francisco?
x1
x2
x3
x4
$8
5
6
4
Capital
Required
(Millions)
$6
3
5
2
Capital available: $10 million
The decisions to be made are listed in the second column of Table 7.1 in the form of yesor-no questions. In each case, giving an answer of yes to the question corresponds to the
decision to make the investment to build the indicated facility (a factory or a warehouse) in
the indicated location (Los Angeles or San Francisco). The capital required for the investment
is given in the rightmost column, where management has made the tentative decision that
the total amount of capital being made available for all the investments is $10 million. (Note
that this amount is inadequate for some of the combinations of investments.) The fourth column shows the estimated net present value (net long-run profit considering the time value of
money) if the corresponding investment is made. (The net present value is 0 if the investment
is not made.) Much of the work of Steve Chan’s management science study (with substantial
help from the president’s staff) goes into developing these estimates of the net present values.
As specified by the company’s president, Armando Ortega, the objective now is to find the
feasible combination of investments that maximizes the total net present value.
Introducing Binary Decision Variables for the Yes-or-No Decisions
As summarized in the second column of Table 7.1, the problem facing management is to make
four interrelated yes-or-no decisions. To formulate a mathematical model for this problem,
Steve Chan needs to introduce a decision variable for each of these decisions. Since each decision has just two alternatives, choose yes or choose no, the corresponding decision variable
only needs to have two values (one for each alternative). Therefore, Steve uses a binary variable, whose only possible values are 0 and 1, where 1 corresponds to the decision to choose yes
and 0 corresponds to choosing no.
These decision variables are shown in the second column of Table 7.2. The final two columns give the interpretation of a value of 1 and 0, respectively.
Dealing with Interrelationships between the Decisions
Recall that management wants no more than one new warehouse to be built. In terms of the
corresponding decision variables, x3 and x4, this means that no more than one of these variables is allowed to have the value 1. Therefore, these variables must satisfy the constraint
​​x​  3​​ + x​ ​  4​​ ≤ 1​
as part of the mathematical model for the problem.
These two alternatives (build a warehouse in Los Angeles or build a warehouse in San
Francisco) are referred to as mutually exclusive alternatives because choosing one of
these alternatives excludes choosing the other. Groups of two or more mutually exclusive
TABLE 7.2 Binary Decision Variables for the California Manufacturing Co. Problem
Decision
Number
Decision
Variable
Possible
Value
Interpretation
of a Value of 1
Interpretation
of a Value of 0
1
2
3
4
x1
x2
x3
x4
0 or 1
0 or 1
0 or 1
0 or 1
Build a factory in Los Angeles
Build a factory in San Francisco
Build a warehouse in Los Angeles
Build a warehouse in San Francisco
Do not build this factory
Do not build this factory
Do not build this warehouse
Do not build this warehouse
7.1 A Case Study: The California Manufacturing Co. Problem 247
With a group of mutually
exclusive alternatives, only
one of the corresponding
binary decision variables
can equal 1.
alternatives arise commonly in BIP problems. For each such group where at most one of the
alternatives can be chosen, the constraint on the corresponding binary decision variables has
the form shown above, namely, the sum of these variables must be less than or equal to 1.
For some groups of mutually exclusive alternatives, management will exclude the possibility
of choosing none of the alternatives, in which case the constraint will set the sum of the corresponding binary decision variables equal to 1.
The California Manufacturing Co. problem also has another important kind of restriction.
Management will allow a warehouse to be built in a particular city only if a factory also is
being built in that city. For example, consider the situation for Los Angeles (LA).
If decide no, do not build a factory in LA (i.e., if choose x1 = 0),
then cannot build a warehouse in LA (i.e., must choose x3 = 0).
If decide yes, do build a factory in LA (i.e., if choose x1 = 1),
then can either build a warehouse in LA or not (i.e., can choose either x3 = 1 or 0).
How can these interrelationships between the factory and warehouse decisions for LA be
expressed in a constraint for a mathematical model? The key is to note that, for either value of
x1, the permissible value or values of x3 are less than or equal to x1. Since x1 and x3 are binary
variables, the constraint
​​x​  3​​ ≤ ​x​  1​​​
forces x3 to take on a permissible value given the value of x1.
Exactly the same reasoning leads to
​​x​  4​​ ≤ ​x​  2​​​
One yes-or-no decision is
contingent on another yesor-no decision if the first
one is allowed to be yes
only if the other one is yes.
as the corresponding constraint for San Francisco. Just as for Los Angeles, this constraint forces
having no warehouse in San Francisco (x4 = 0) if a factory will not be built there (x2 = 0),
whereas going ahead with the factory there (x2 = 1) leaves open the decision to build the warehouse there (x4 = 0 or 1).
For either city, the warehouse decision is referred to as a contingent decision, because
the decision depends on a prior decision regarding whether to build a factory there. In general,
one yes-or-no decision is said to be contingent on another yes-or-no decision if it is allowed to
be yes only if the other is yes. As above, the mathematical constraint expressing this relationship requires that the binary variable for the former decision must be less than or equal to the
binary variable for the latter decision.
The rightmost column of Table 7.1 reveals one more interrelationship between the four
decisions, namely, that the amount of capital expended on the four facilities under consideration cannot exceed the amount available ($10 million). Therefore, the model needs to include
a constraint that requires
​Capital expended ≤ $10 million​
Excel Tip: Beware that
rounding errors can occur
with Excel. Therefore, even
when you add a constraint
that a changing cell has to
be binary, Excel occasionally will return a noninteger
value very close to an
integer (e.g., 1.23E-10,
meaning 0.000000000123).
When this happens, you
can replace the noninteger
value with the proper
integer value.
How can the amount of capital expended be expressed in terms of the four binary decision
variables? To start this process, consider the first yes-or-no decision (build a factory in Los
Angeles?). Combining the information in the rightmost column of Table 7.1 and the first row
of Table 7.2,
$6 million  if x​ ​  1​​ = 1
Capital expended on factory in Los Angeles = ​​
​​​ ​
​
​​​
{
if x​ ​  1 ​​
​​ = 0​
​​​
​  ​  0
​
= $6 million times ​x​  1​​
By the same reasoning, the amount of capital expended on the other three investment opportunities (in units of millions of dollars) is 3x2, 5x3, and 2x4, respectively. Consequently,
​​Capital expended = 6​x​  1​​ + 3​x​  2​​ + 5​x​  3​​ + 2​x​  4​​​
(in millions of dollars)​​
Therefore, the constraint becomes
​6​x​  1​​ + 3​x​  2​​ + 5​x​  3​​ + 2​x​  4​​  ≤ 10​
248 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
Excel Tip: In the Solver
Options, the Integer
Optimality (%) setting
(1 ­percent by default)
causes Solver to stop solving an integer programming
problem when it finds a feasible solution whose objective function value is within
the specified percentage of
being optimal. (In ­Analytic
Solver, this option is called
Integer Tolerance and is
found in the Engine tab
of the Model pane.) This
is useful for very large
BIP problems since it may
enable finding a nearoptimal solution when
finding an optimal solution
in a reasonable period of
time is not possible. For
smaller problems (e.g., all
the problems in this book),
this option should be set to
0 to guarantee finding an
optimal solution.
Note how helpful the range
names are for interpreting
this BIP spreadsheet model.
The BIP Model
As indicated by Armando Ortega in his conversation with Steve Chan, management’s objective is to find the feasible combination of investments that maximizes the total net present
value of these investments. Thus, the value of the objective function should be
​NPV = Total net present value​
If the investment is made to build a particular facility (so that the corresponding decision
variable has a value of 1), the estimated net present value from that investment is given in the
fourth column of Table 7.1. If the investment is not made (so the decision variable equals 0),
the net present value is 0. Therefore, continuing to use units of millions of dollars,
​NPV = 8​x​  1​​ + 5​x​  2​​ + 6​x​  3​​ + 4​x​  4​​​
is the quantity to enter into the objective cell to be maximized.
Incorporating the constraints developed in the preceding subsection, the complete BIP
model then is shown in Figure 7.1. The format is basically the same as for linear programming
models. The one key difference arises when using Solver. Each of the decision variables (cells
C18:D18 and C16:D16) is constrained to be binary. In Excel’s Solver, this is accomplished
in the Add Constraint dialog box by choosing each range of changing cells as the left-hand
side and then choosing bin from the pop-up menu. In Analytic Solver, this is accomplished by
selecting each range of changing cells, and then under the Constraints menu on the Analytic
Solver ribbon, choose Binary in the Variable Type>Bound submenu. The other constraints
shown in Solver (see the lower left-hand side of Figure 7.1) have been made quite intuitive
by using the suggestive range names given in the lower right-hand side of the figure. For convenience, the equations entered into the output cells in E12 and D20 use a SUMPRODUCT
function that includes C17:D17 and either C11:D11 or C5:D5 because the blanks or ≤ signs
in these rows are interpreted as zeroes by Solver.
Solver gives the optimal solution shown in C18:D18 and C16:D16 of the spreadsheet,
namely, build factories in both Los Angeles and San Francisco, but do not build any warehouses. The objective cell (D20) indicates that the total net present value from building these
two factories is estimated to be $13 million.
Performing What-If Analysis
Solver’s sensitivity report
is not available for integer
programming problems.
Trial-and-error and/or a
parameter analysis report
can be used to perform
what-if analysis for integer
programming problems.
Now that Steve Chan has used the BIP model to determine what should be done when the
amount of capital being made available to these investments is $10 million, his next task is to
perform what-if analysis on this amount. Recall that Armando Ortega wants him to determine
what the effect would be if this amount were changed to anything else between $5 million and
$15 million.
In Chapter 5, we described three different methods of performing what-if analysis on a
linear programming spreadsheet model when there is a change in a constraint: using trial
and error with the spreadsheet, generating a parameter analysis report, or referring to Solver’s sensitivity report. The first two of these can be used on integer programming problems
in exactly the same way as for linear programming problems. The third method, however,
does not work. The sensitivity report is not available for integer programming problems.
This is because the concept of a shadow price and allowable range no longer applies. In
contrast to linear programming, the objective function values for an integer programming
problem do not change in a predictable manner when the right-hand side of a constraint is
changed.
It is straightforward to determine the impact of changing the amount of available capital by
trial and error. Simply try different values in the data cell CapitalAvailable (G12) and re-solve
with Solver. However, a more systematic way to perform this analysis is to generate a parameter analysis report using Analytic Solver. The parameter analysis report works for integer
programming models in exactly the same way as it does for linear programming models (as
described in Section 5.3 in the subsection entitled Using a Parameter Analysis Report Generated by Analytic Solver to Do Sensitivity Analysis Systematically).
7.1 A Case Study: The California Manufacturing Co. Problem 249
FIGURE 7.1
A spreadsheet formulation
of the BIP model for
the California Manufacturing Co. case study
where the changing cells
BuildFactory? (C18:D18)
and BuildWarehouse?
(C16:D16) give the
optimal solution obtained
by Solver.
A
1
2
B
C
D
E
F
G
California Manufacturing Co. Facility Location Problem
3
NPV ($millions)
LA
SF
4
Warehouse
6
4
Factory
8
5
5
6
7
8
Capital Required
9
($millions)
LA
SF
10
Warehouse
5
2
Factory
6
3
Total
Maximum
15
Build?
LA
SF
Warehouses
Warehouses
16
Warehouse
0
0
0
<=
<=
1
1
11
12
Capital
Capital
Spent
Available
<=
9
10
13
14
17
18
Factory
<=
1
19
20
Total NPV ($millions)
Solver Parameters
Set Objective Cell: TotalNPV
To: Max
By Changing Variable Cells:
BuildWarehouse?, BuildFactory?
Subject to the Constraints:
BuildFactory? = binary
BuildWarehouse? = binary
BuildWarehouse? <= BuildFactory?
CapitalSpent <= CapitalAvailable
TotalWarehouses <= MaxWarehouses
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
$13
E
10
Capital
11
Spent
12
= SUMPRODUCT(CapitalRequired,Build?)
13
14
Total
15
Warehouses
= SUM(BuildWarehouse?)
16
C
20
Range Name
Cells
Build?
BuildWarehouse?
BuildFactory?
CapitalAvailable
CapitalRequired
CapitalSpent
MaxWarehouses
NPV
TotalNPV
TotalWarehouses
C16:D18
C16:D16
C18:D18
G12
C10:D12
E12
G16
C4:D6
D20
E16
D
Total NPV ($millions) =SUMPRODUCT(NPV,Build?)
After defining CapitalAvailable (G12) as a parameter cell with values ranging from 5 to 15
($millions), the parameter analysis report shown in Figure 7.2 was generated by executing the
series of steps outlined in Section 5.3. Note how Figure 7.2 shows the effect on the optimal
solution and the resulting total net present value of varying the amount of capital being made
available.
250 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
FIGURE 7.2
The parameter analysis
report generated by
Analytic Solver that
shows the effect on the
optimal solution and the
resulting total net present
value of systematically
varying the amount of
capital being made available for these investments.
A
1 CapitalAvailable
2
5
3
6
7
4
8
5
9
6
10
7
11
8
12
9
13
10
14
11
12
15
B
BuildWarehouseLA?
0
0
0
0
0
0
0
0
0
1
1
C
BuildWarehouseSF?
1
1
1
1
0
0
1
1
1
0
0
D
BuildFactoryLA?
0
0
0
0
1
1
1
1
1
1
1
E
BuildFactorySF?
1
1
1
1
1
1
1
1
1
1
1
F
TotalINPV
9
9
9
9
13
13
17
17
17
19
19
What-if analysis also could be performed on any of the other data cells—NPV (C4:D6),
CapitalRequired (C10:D12), and MaxWarehouses (G16)—in a similar way with a parameter
analysis report (or by using trial and error with the spreadsheet). However, a careful job was
done in developing good estimates of the net present value of each of the possible investments, and there is little uncertainty in the values entered in the other data cells, so Steve
Chan decides that further what-if analysis is not needed.
Management’s Conclusion
Steve Chan’s report is delivered to Armando Ortega within the two-week deadline. The report
recommends the plan presented in Figure 7.1 (build a factory in both Los Angeles and San
Francisco but no warehouses) if management decides to stick with its tentative decision to make
$10 million of capital available for these investments. One advantage of this plan is that it only
uses $9 million of this capital, which frees up $1 million of capital for other project proposals
currently being investigated. The report also highlights the results shown in F
­ igure 7.2 while
emphasizing two points. One is that a heavy penalty would be paid (a reduction in the total net
present value from $13 million to $9 million) if the amount of capital being made available
were to be reduced below $9 million. The other is that increasing the amount of capital being
made available by just $1 million (from $10 million to $11 million) would enable a substantial
increase of $4 million in the total net present value (from $13 million to $17 million). However,
a much larger further increase in the amount of capital being made available (from $11 million
to $14 million) would be needed to enable a considerably smaller further increase in the total net
present value (from $17 million to $19 million).
Armando Ortega deliberates with other members of top management before making a
decision. It is quickly concluded that increasing the amount of capital being made available all
the way up to $14 million would be stretching the company’s financial resources too dangerously to justify the relatively small payoff. However, there is considerable discussion of the
pros and cons of the two options of using either $9 million or $11 million of capital. Because
of the large payoff from the latter option (an additional $4 million in total net present value),
management finally decides to adopt the plan presented in row 8 of Figure 7.2. Thus, the
company will build new factories in both Los Angeles and San Francisco as well as a new
warehouse in San Francisco, with an estimated total net present value of $17 million. However, because of the large capital requirements of this plan, management also decides to defer
building the warehouse until the two factories are completed so that their profits can help
finance the construction of the warehouse.
Review
Questions
1. What are the four interrelated decisions that need to be made by the management of the
­California Manufacturing Co.?
2. Why are binary decision variables appropriate to represent these decisions?
3. What is the objective specified by management for this problem?
4. What are the mutually exclusive alternatives in this problem? What is the form of the resulting
constraint in the BIP model?
5. What are the contingent decisions in this problem? For each one, what is the form of the
resulting constraint in the BIP model?
6. What is the tentative managerial decision on which what-if analysis needs to be performed?
7.2 Using BIP for Project Selection: The Tazer Corp. Problem 251
7.2
USING BIP FOR PROJECT SELECTION: THE TAZER CORP. PROBLEM
The California Manufacturing Co. case study focused on four proposed projects: (1) build
a factory in Los Angeles, (2) build a factory in San Francisco, (3) build a warehouse in Los
Angeles, and (4) build a warehouse in San Francisco. Management needed to make yes-or-no
decisions about which of these projects to select. This is typical of many applications of BIP.
However, the nature of the projects may vary considerably from one application to the next.
Instead of the proposed construction projects in the case study, our next example involves the
selection of research and development projects.
This example is adapted from both Case 3-7 and its continuation in Section 13.8, but all the
relevant information is repeated below.
The Tazer Corp. Problem
Tazer Corp., a pharmaceutical manufacturing company, is beginning the search for a new
breakthrough drug. The following five potential research and development projects have been
identified for attempting to develop such a drug.
Project Up:Develop a more effective antidepressant that does not cause serious
mood swings.
Project Stable:
Develop a drug that addresses manic depression.
Project Choice:
Develop a less intrusive birth control method for women.
Project Hope:
Develop a vaccine to prevent HIV infection.
Project Release: Develop a more effective drug to lower blood pressure.
The objective is to choose
the projects that will maximize the expected profit
while satisfying the budget
constraint.
In contrast to Case 3-7, Tazer management now has concluded that the company cannot
devote enough money to research and development to undertake all of these projects. Only
$1.2 billion is available, which will be enough for only two or three of the projects. The first row
of Table 7.3 shows the amount needed (in millions of dollars) for each of these projects, where
the footnote points out the budget constraint that only $1.2 billion is available. The second row
estimates each project’s probability of being successful. If a project is successful, it is estimated
that the resulting drug would generate the revenue shown in the third row. Thus, the expected
revenue (in the statistical sense) from a potential drug is the product of its numbers in the second
and third rows, whereas its expected profit is this expected revenue minus the investment given
in the first row. These expected profits are shown in the bottom row of Table 7.3.
Tazer management now wants to determine which of these projects should be undertaken
to maximize their expected total profit.
Formulation with Binary Variables
Because the decision for each of the five proposed research and development projects is a yesor-no decision, the corresponding decision variables are binary variables. Thus, the decision
variable for each project has the following interpretation.
1, if approve the project
​​Decision Variable​  =​  ​​
​​​ ​
​​​​​
{0, if reject the project
Let x1, x2, x3, x4, and x5 denote the decision variables for the respective projects in the order in
which they are listed in Table 7.3.
TABLE 7.3
Data for the Tazer
Project Selection
Problem
Project
R&D investment ($million)*
Success rate
Revenue if successful ($million)
Expected profit ($million)
1 (Up)
2 (Stable)
3 (Choice)
4 (Hope)
5 (Release)
400
50%
1,400
300
300
35%
1,200
120
600
35%
2,200
170
500
20%
3,000
100
200
45%
600
70
* Only $1.2 billion is available for R&D investment.
An Application Vignette
The Midcontinent Independent System Operator, Inc. (MISO)
is a nonprofit organization formed in 1998 (and renamed in
2013) to administer the generation and transmission of electricity throughout the midwestern United States. It serves approximately 40 million customers (both individuals and businesses)
through its control of approximately 60,000 miles of highvoltage transmission lines and more than 1,000 power plants
capable of generating massive amounts of of electricity. This
infrastructure spans many midwestern and southern states plus
the Canadian province of Manitoba.
The key mission of any regional transmission organization
is to reliably and efficiently provide the electricity needed by its
customers. MISO transformed the way this was done by using
mixed binary integer programming to minimize the total cost
of providing the needed electricity. Each main binary variable
in the model represents a yes-or-no decision about whether
a particular power plant should be on during a particular time
period. After solving this model, the results are then fed into a
linear programming model to set electricity output levels and
establish prices for electricity trades.
The mixed BIP model is a massive one which began with
about 3,300,000 continuous variables, 450,000 binary variables, and 3,900,000 functional constraints. A special technique (Lagrangian relaxation) is used to solve such a huge
model.
This innovative application of management science yielded
savings of approximately $2.5 billion over the four years from
2007 to 2010, with an additional savings of about $7 billion
expected through 2020. These dramatic results led to MISO
winning the prestigious first prize in the 2011 international competition for the Franz Edelman Award for Achievement in Operations Research and the Management Sciences.
Source: B. Carlson and 12 co-authors, “MISO Unlocks Billions in Savings
through the Application of Operations Research for Energy and Ancillary
Services Markets,” Interfaces 42, no. 1 (January–February 2012), pp. 58–73.
(A link to this article is provided at www.mhhe.com/Hillier6e.)
If a project is rejected, there is neither any profit nor any loss, whereas the expected profit
if a project is approved is given in the bottom row of Table 7.3. Therefore, when using units of
millions of dollars, the expected total profit is
​P = 300​x​  1​​ + 120​x​  2​​ + 170​x​  3​​ + 100​x​  4​​ + 70​x​  5​​​
The objective is to select the projects to approve that will maximize this expected total profit
while satisfying the budget constraint.
Other than requiring the decision variables to be binary, the budget constraint limiting the
total investment to no more than $1.2 billion is the only constraint that has been imposed by
Tazer management on the selection of these research and development projects.
Referring to the first row of Table 7.3, this constraint can be expressed in terms of the decision variables as
​400​x​  1​​ + 300​x​  2​​ + 600​x​  3​​ + 500​x​  4​​ + 200​x​  5​​ ≤ 1,200​
With this background, the stage now is set for formulating a BIP spreadsheet model for this
problem.
A BIP Spreadsheet Model for the Tazer Problem
Figure 7.3 shows a BIP spreadsheet model for this problem. The data in Table 7.3 have been
transferred into cells C5:G8. The changing cells are DoProject? (C10:G10) and the objective
cell is TotalExpectedProfit (H8). The one functional constraint is depicted in cells H5:J5. In
addition, the changing cells DoProject? are constrained to be binary, as shown in the Solver
Parameters box.
The changing cells DoProject? (C10:G10) in Figure 7.3 show the optimal solution that has
been obtained by Solver, namely,
Choose Project Up, Project Choice, and Project Release.
The objective cell indicates that the resulting total expected profit is $540 million.
Review
Questions
252
1. How are binary variables used to represent managerial decisions on which projects from a
group of proposed projects should be selected for approval?
2. What types of projects are under consideration in the Tazer Corp. problem?
3. What is the objective for this problem?
7.3 Using BIP for The Selection of Sites for Emergency Services Facilities: The Caliente City Problem 253
FIGURE 7.3
A spreadsheet formulation of the BIP model for the Tazer Corp. project selection problem where the changing cells DoProject?
(C10:G10) give the optimal solution obtained by Solver.
A
1
2
3
4
5
6
7
8
9
10
8
B
C
D
E
F
G
H
I
J
Stable
300
35%
1,200
120
Choice
600
35%
2,200
170
Hope
500
20%
3,000
100
Release
200
45%
600
70
Total
1,200
<=
Budget
1,200
0
1
0
1
Tazer Corp. Project Selection Problem
R&D Investment ($million)
Success Rate
Revenue If Successful ($million)
Expected Profit ($million)
Do Project?
B
Expected Profit ($million)
Up
400
50%
1,400
300
1
C
=C7*C6-C5
Range Name
Cells
Budget
J5
DoProject?
C10:G10
ExpectedProfit
C8:G8
RandDInvestment
Revenue
C5:G5
C7:G7
4
SuccessRate
C6:G6
6
TotalExpectedProfit
TotalRandD
H8
H5
7
D
=D7*D6-D5
E
=E7*E6-E5
540
F
=F7*F6-F5
G
=G7*G6-G5
H
5
8
Total
=SUMPRODUCT(RandDInvestment,DoProject?)
=SUMPRODUCT(ExpectedProfit,DoProject?)
Solver Parameters
Set Objective Cell: TotalExpectedProfit
To: Max
By Changing Variable Cells:
DoProject?
Subject to the Constraints:
DoProject? = binary
TotalRandD <= Budget
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
7.3 USING BIP FOR THE SELECTION OF SITES FOR EMERGENCY
SERVICES FACILITIES: THE CALIENTE CITY PROBLEM
Although the problem encountered in the California Manufacturing Co. case study can be
described as a project selection problem (as was done at the beginning of the preceding section), it could just as well have been called a site selection problem. Recall that the company’s
management needed to select a site (Los Angeles or San Francisco) for its new factory as
well as for its possible new warehouse. For either of the possible sites for the new factory (or
the warehouse), there is a yes-or-no decision for whether that site should be selected, so it
becomes natural to represent each such decision by a binary decision variable.
Various kinds of site selection problems are one of the most common types of applications
of BIP. The kinds of facilities for which sites need to be selected can be of any type. In some
cases, several sites are to be selected for several facilities of a particular type, whereas only a
single site is to be selected in other cases.
254 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
We will focus here on the selection of sites for emergency services facilities. These facilities might be fire stations, police stations, ambulance centers, and so forth. In any of these
cases, the overriding concern commonly is to provide facilities close enough to each part of
the area being served that the response time to an emergency anywhere in the area will be sufficiently small. The form of the BIP model then will be basically the same regardless of the
specific type of emergency services being considered.
To illustrate, let us consider an example where sites are being selected for fire stations. For
simplicity, this example will divide the area being served into only eight tracts instead of the
many dozens or hundreds that would be typical in real applications.
The Caliente City Problem
Caliente City is located in a particularly warm and arid part of the United States, so it is
especially prone to the occurrence of fires. The city has become a popular place for senior
citizens to move to after retirement, so it has been growing rapidly and spreading well beyond
its original borders. However, the city still has only one fire station, located in the congested
center of the original town site. The result has been some long delays in fire trucks reaching
fires in the outer parts of the city, causing much more damage than would have occurred with
a prompt response. The city’s residents are very unhappy about this, so the city council has
directed the city manager to develop a plan for locating multiple fire stations throughout the
city (including perhaps moving the current fire station) that would greatly reduce the response
time to any fire. In particular, the city council has adopted the following policy about the
maximum acceptable response time for fire trucks to reach a fire anywhere in the city after
being notified about the fire.
​Response time ≤ 10 minutes​
The objective is to
minimize the total cost of
ensuring a response time of
no more than 10 minutes.
Having had a management science course in college, the city manager recognizes that BIP
provides her with a powerful tool for analyzing this problem. To get started, she divides the
city into eight tracts and then gathers data on the estimated response time for a fire in each
tract from a potential fire station in each of the eight tracts. These data are shown in Table 7.4.
For example, if a decision were to be made to locate a fire station in tract 1 and if that fire station were to be used to respond to a fire in any of the tracts, the second column of Table 7.4
shows what the (estimated) response time would be. (Since the response time would exceed 10
­minutes for a fire in tracts 3, 5, 6, 7, or 8, a fire station actually would need to be located nearer
to each of these tracts to satisfy the city council’s new policy.) The bottom row of Table 7.4
shows what the cost would be of acquiring the land and constructing a fire station in any of the
eight tracts. (The cost is far less for tract 5 because the current fire station already is there so
only a modest renovation is needed if the decision is made to retain a fire station there.)
The objective now is to determine which tracts should receive a fire station to minimize the
total cost of the stations while ensuring that each tract has at least one station close enough to
respond to a fire in no more than 10 minutes.
TABLE 7.4 Response Time and Cost Data for the Caliente City Problem
Fire Station in Tract
Response
times
(minutes)
for a fire
in tract
Cost of station
($thousands)
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
2
9
17
10
21
25
14
30
8
3
8
13
12
15
22
24
18
10
4
19
16
7
18
15
9
12
20
2
13
21
7
14
23
16
21
18
5
15
13
17
22
14
8
21
11
3
15
9
16
21
22
6
9
14
2
8
28
25
17
12
12
8
9
3
350
250
450
300
50
400
300
200
7.3 Using BIP for The Selection of Sites for Emergency Services Facilities: The Caliente City Problem 255
Formulation with Binary Variables
For each of the eight tracts, there is a yes-or-no decision as to whether that tract should receive
a fire station. Therefore, we let x1, x2, . . . , x8 denote the corresponding binary decision variables, where
1, if tract j is selected to receive a fire station
​​x​  j​​ = ​​
​​​​
{​​​0,​  if not
for j = 1, 2, . . . , 8.
Since the objective is to minimize the total cost of the fire stations that will satisfy the city
council’s new policy on response times, the total cost needs to be expressed in terms of these
decision variables. Using units of thousands of dollars while referring to the bottom row of
Table 7.4, the total cost is
​C = 350​x​  1​​ + 250​x​  2​​ + 450​x​  3​​ + 300​x​  4​​ + 50​x​  5​​ + 400​x​  6​​ + 300​x​  7​​ + 200​x​  8​​​
We also need to formulate constraints in terms of these decision variables that will ensure
that no response times exceed 10 minutes. For example, consider tract 1. When a fire occurs
there, the row for tract 1 in Table 7.4 indicates that the only tracts close enough that a fire
station would provide a response time not exceeding 10 minutes are tract 1 itself, tract 2, and
tract 4. Thus, at least one of these three tracts needs to have a fire station. This requirement is
expressed in the constraint
​​x​  1​​ + ​x​  2​​ + ​x​  4​​ ≥ 1​
This constraint ensures that
the response time for a fire
in tract 1 will be no more
than 10 minutes.
Incidentally, this constraint is called a set covering constraint because it covers the
requirement of having a fire station located in at least one member of the set of tracts (tracts
1, 2, and 4) that are within 10 minutes of tract 1. In general, any constraint where a sum of
binary variables is required to be greater than or equal to one is referred to as a set covering
constraint.
Applying the above reasoning for tract 1 to all the tracts leads to the following constraints.
Tract 1: ​x​  1​​ +​x​  2​​ ​
+​x​  4​​ ​
​
​
​
≥ 1
Tract 2: ​x​  1​​ +​x​  2​​ +​x​  3​​ ​
​
​
​
​
≥ 1
Tract 3: ​
​x​  2​​ +​x​  3​​ ​
​
+​x​  6​​ ​
​
≥ 1
Tract 4: ​x​  1​​ ​
​
+​x​  ​​ ​
​
+​x​  ​​ ​
≥ 1
​​
​  ​  ​  ​  ​  ​  ​  ​  ​  ​  ​  ​  4​  ​
​  ​
​  ​  7​  ​
​
​​​
Tract 5: ​ ​
​
​
+​x​  5​​ ​
+​x​  7​​ ​
≥ 1
Tract 6: ​ ​
​x​  3​​
​
+​x​  6​​ ​
+​x​  8​​ ≥ 1
Tract 7: ​ ​
​
​x​  4​​ ​
​
+​x​  7​​ +​x​  8​​ ≥ 1
Tract 8: ​ ​
​
​
​
+​x​  6​​ +​x​  7​​ +​x​  8​​ ≥ 1
These set covering constraints (along with requiring the variables to be binary) are all that is
needed to ensure that each tract has at least one fire station close enough to respond to a fire
in no more than 10 minutes.
This type of BIP model (minimizing total cost where all the functional constraints are
set covering constraints) is called a set covering problem. Such problems arise fairly frequently. In fact, you will see another example of a set covering problem in Section 7.4.
Having identified the nature of the constraints for the Caliente City problem, it now is
fairly straightforward to formulate its BIP spreadsheet model.
A BIP Spreadsheet Model for the Caliente City Problem
Figure 7.4 shows a BIP spreadsheet model for this problem. The data cells ResponseTime
(D5:K12) show all the response times given in Table 7.4 and CostOfStation (D14:K14) provides the cost data from the bottom row of this table. There is a yes-or-no decision for each tract
as to whether a fire station should be located there, so the changing cells are StationInTract?
(D29:K29). The objective is to minimize total cost, so the objective cell is TotalCost (N29).
The set covering constraints are displayed in cells L17:N24. The Station-InTract? changing
cells have been constrained to be binary, as can be seen in the Solver Parameters box.
256 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
FIGURE 7.4
A spreadsheet formulation of the BIP model for the Caliente City site selection problem where the changing cells StationInTract?
(D29:K29) show the optimal solution obtained by Solver.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
15
16
17
18
19
B
C
D
E
F
G
H
I
J
K
1
2
3
4
5
6
7
8
1
2
9
17
10
21
25
14
30
2
8
3
8
13
12
15
22
24
6
22
14
8
21
11
3
15
9
7
16
21
22
6
9
14
2
8
8
28
25
17
12
12
8
9
3
350
250
450
300
50
400
300
200
1
1
0
1
0
0
0
0
1
1
1
0
0
0
0
0
0
1
1
0
0
1
0
0
1
0
0
1
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
1
0
1
0
0
0
1
1
0
1
1
0
0
0
0
0
1
1
1
1
0
Fire Station in Tract
2
3
4
5
6
1
0
0
0
0
Caliente City Fire Station Location Problem
Response
Times
(minutes)
for a Fire
in Tract
Cost of Station
($thousands)
Response
Time
<=
10
Minutes?
1
2
3
4
5
6
7
8
Station in Tract?
Fire Station in Tract
3
4
5
18
9
23
10
12
16
4
20
21
19
2
18
16
13
5
7
21
15
18
7
13
15
14
17
7
1
L
M
N
Number
Covering
1
1
1
1
1
1
2
2
>=
>=
>=
>=
>=
>=
>=
>=
1
1
1
1
1
1
1
1
8
1
J
K
=IF(J5<=MaxResponseTime,1,0)
=IF(J6<=MaxResponseTime,1,0)
=IF(J7<=MaxResponseTime,1,0)
=IF(K5<=MaxResponseTime,1,0)
=IF(K6<=MaxResponseTime,1,0)
=IF(K7<=MaxResponseTime,1,0)
Solver Parameters
Set Objective Cell: TotalCost
To: Min
By Changing Variable Cells:
StationInTract?
Subject to the Constraints:
StationInTract? = binary
NumberCovering >= One
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
Total
Cost
($thousands)
750
L
Number
Covering
=SUMPRODUCT(D17:K17,StationInTract?)
=SUMPRODUCT(D18:K18,StationInTract?)
=SUMPRODUCT(D19:K19,StationInTract?)
Range Name
Cells
CostOfStation
D14:K14
MaxResponseTime
B21
NumberCovering
L17:L24
One
ResponseTime
N17:N24
D5:K12
StationInTract?
D29:K29
TotalCost
N29
N
26
Total
27
Cost
28
($thousands)
29
=SUMPRODUCT(CostOfStation,StationInTract?)
7.4 Using BIP for Crew Scheduling: The Southwestern Airways Problem 257
After running Solver, the optimal solution shown in the changing cells StationInTract?
(D29:K29) in Figure 7.4 is obtained, namely,
Select tracts 2, 7, and 8 as the sites for fire stations.
The objective cell TotalCost (N29) indicates that the resulting total cost is $750,000.
Review
Questions
1. How are binary variables used to represent managerial decisions regarding which site or sites
should be selected for new facilities?
2. What are some types of emergency services facilities for which sites may need to be selected?
3. What was the objective for the Caliente City problem?
4. What is a set covering constraint and what is a set covering problem?
7.4 USING BIP FOR CREW SCHEDULING: THE SOUTHWESTERN
AIRWAYS PROBLEM
The travel industry includes airlines, rail travel, cruise ships, tour companies, etc. All members of this industry face the same problem of needing to assign their crews to the various
trips being offered in order to serve the customers on those trips. When assigning a crew to a
sequence of trips, it would be particularly convenient to have the last trip end where the first
trip begins. With many crews and numerous trips to be covered, what is the most efficient way
of assigning all the crews to all the trips? This is referred to as the crew scheduling problem.
One way of formulating the crew scheduling problem is to identify many feasible overlapping sequences of trips for each crew. The objective then is to assign each crew to a sequence
of trips so as to cover all the trips at a minimum cost. Thus, for each feasible sequence of trips,
there is a yes-or-no decision as to whether a crew should be assigned to that sequence, so a
binary decision variable can be used to represent that decision.
For many years, airline companies have been using BIP models to determine how to do
their crew scheduling in the most cost-efficient way. Some airlines have saved many millions
of dollars annually through this application of BIP. Consequently, other segments of the travel
industry now are also using BIP in this way. For example, the application vignette in this section describes how Netherlands Railways achieved a dramatic increase in profits by applying
BIP (and related techniques) in a variety of ways, including crew scheduling.
To illustrate the approach, consider the following miniature example of airline crew
scheduling.
The Southwestern Airways Problem
Southwestern Airways needs to assign its crews to cover all its upcoming flights. We will
focus on just one portion of this problem, namely, the problem of assigning three crews based
in San Francisco (SFO) to the 11 flights shown in Figure 7.5. Thus, each crew needs to be
assigned to a sequence of flights that begins and ends in San Francisco. Furthermore, the different sequences of flights for these three crews need to “cover” (i.e., include) all 11 flights.
These 11 flights are listed in the first column of Table 7.5. The other 12 columns show
the 12 feasible sequences of flights for a crew. The numbers in each column indicate the
order of the flights. For example, sequence 4 says for one crew to begin with flight 1 out of
San Francisco, then take flight 4, then take flight 6, and then finally take flight 8 back to San
Francisco. This sequence thereby covers 4 of the 11 flights, so the other two sequences for the
other two crews would need to cover the remaining 7 flights.
The key requirement is that (at most) three of the sequences need to be chosen (one per
crew) in such a way that every flight is covered. (It is permissible to have more than one crew
on a flight, where the extra crews would fly as passengers, but union contracts require that
the extra crews still be paid for their time as if they were working.) The cost of assigning a
crew to a particular sequence of flights is given (in thousands of dollars) in the bottom row
of the table. The objective is to minimize the total cost of the crew assignments that cover
all the flights.
258 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
FIGURE 7.5
The arrows show the
11 Southwestern Airways
flights that need to be
covered by the three crews
based in San Francisco.
Seattle
(SEA)
San Francisco
(SFO)
Chicago
(ORD)
Denver
(DEN)
Los Angeles
(LAX)
TABLE 7.5 Data for the Southwestern Airways Problem
Feasible Sequence of Flights
Flight
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
1
San Francisco to Los Angeles
(SFO–LAX)
San Francisco to Denver
(SFO–DEN)
San Francisco to Seattle
(SFO–SEA)
Los Angeles to Chicago
(LAX–ORD)
Los Angeles to San Francisco
(LAX–SFO)
Chicago to Denver
(ORD–DEN)
Chicago to Seattle
(ORD–SEA)
Denver to San Francisco
(DEN–SFO)
Denver to Chicago
(DEN–ORD)
Seattle to San Francisco
(SEA–SFO)
Seattle to Los Angeles
(SEA–LAX)
Cost, $1,000s
2
3
1
4
5
6
1
1
7
8
1
1
1
1
2
3
3
4
3
4
4
7
5
3
3
4
5
7
2
4
2
6
3
5
2
2
4
2
5
2
3
1
4
3
2
12
1
3
3
11
1
1
2
10
1
2
2
9
8
5
2
4
4
2
9
9
8
9
Formulation with Binary Variables
With 12 feasible sequences of flights, we have 12 yes-or-no decisions:
​​Should sequence j be assigned to a crew?​
(j = 1, 2, . . . , 12)​​
Therefore, we use 12 binary variables to represent these respective decisions:
1, if sequence j is assigned to a crew
​​x​  j​​ = ​​ ​​​
{0,​  otherwise ​​​​
Since the objective is to minimize the total cost of the three crew assignments, we now
need to express the total cost in terms of these binary decision variables. Referring to the bottom row of Table 7.5, this total cost (in units of thousands of dollars) is
​C = 2​x​  1​​ + 3​x​  2​​ + 4​x​  3​​ + 6​x​  4​​ + 7​x​  5​​ + 5​x​  6​​ + 7​x​  7​​ + 8​x​  8​​ + 9​x​  9​​ + 9​x​  10​​ + 8​x​  11​​ + 9​x​  12​​​
An Application Vignette
Netherlands Railways (Nederlandse Spoorwegen Reizigers) is the
main Dutch railway operator of passenger trains. In this densely
populated country, about 5,500 passenger trains currently transport approximately 1.1 million passengers on an average workday.
The company’s operating revenues are approximately 1.5 billion
euros (approximately $2 billion) per year.
The amount of passenger transport on the Dutch railway
network has steadily increased over the years, so a national
study in 2002 concluded that three major infrastructure extensions should be undertaken. As a result, a new national timetable for the Dutch railway system, specifying the planned
departure and arrival times of every train at every station,
would need to be developed. Therefore, the management of
Netherlands Railways directed that an extensive management
science study should be conducted over the next few years
to develop an optimal overall plan for both the new timetable
and the usage of the available resources (rolling-stock units
and train crews) for meeting this timetable. A task force consisting of several members of the company’s Department of
Logistics and several prominent management science scholars
from European universities or a software company was formed
to conduct this study.
The new timetable was launched in December 2006, along
with a new system for scheduling the allocation of rolling-stock
units (various kinds of passenger cars and other train units) to
the trains meeting this timetable. A new system also was implemented for scheduling the assignment of crews (with a driver
and a number of conductors in each crew) to the trains. Binary
integer programming and related techniques were used to do
all of this. For example, the BIP model used for crew scheduling closely resembles (except for its vastly larger size) the one
shown in this section for the Southwestern Airlines problem.
This application of management science immediately resulted
in an additional annual profit of approximately $60 million for
the company and this additional profit is expected to increase
to $105 million annually in the coming years. These dramatic
results led to Netherlands Railways winning the prestigious
First Prize in the 2008 international competition for the Franz
Edelman Award for Achievement in Operations Research and
the Management Sciences.
Source: L. Kroon, D. Huisman, E. Abbink, P.-J. Fioole, M. Fischetti, G. Maroti,
A. Schrijver, A. Steenbeck, and R. Ybema, “The New Dutch Timetable:
The OR Revolution,” Interfaces 39, no. 1 (January–February 2009),
pp. 6–17. (A link to this article is provided at www.mhhe.com/Hillier6e.)
With only three crews available to cover the flights, we also need the constraint
​​x​  1​​ + ​x​  2​​ + . . . + ​x​  12​​ ≤ 3​
The most interesting part of this formulation is the nature of each constraint that ensures
that a corresponding flight is covered. For example, consider the last flight in Table 7.5 (Seattle
to Los Angeles). Five sequences (namely, sequences 6, 9, 10, 11, and 12) include this flight.
Therefore, at least one of these five sequences must be chosen. The resulting constraint is
​​x​  6​​ + x​ ​  9​​ + ​x​  10​​ + ​x​  11​​ + ​x​  12​​ ≥ 1​
For each of the 11 flights, the constraint that ensures that the flight is covered is constructed in the same way from Table 7.5 by requiring that at least one of the flight sequences
that includes that flight is assigned to a crew. Thus, 11 constraints of the following form are
needed.
These are set covering
constraints, just like
the constraints in the
Caliente City problem in
Section 7.3.
Flight 1: ​x​  1​​ + ​x​  4​​ + x​ ​  7​​ + ​x​  10​​ ≥ 1
Flight 2: ​x​  2​​ + ​x​  5​​ + x​ ​  8​​ + ​x​  11​​ ≥ 1
​
⋅​
​ ​​
​  ​
​ ​​
​
⋅
​
⋅
Flight 11: ​x​  6​​ + ​x​  9​​ + ​x​  10​​ + ​x​  11​​ + x​ ​  12​​ ≥ 1
Note that these constraints have the same form as the constraints for the Caliente City
problem in Section 7.3 (a sum of certain binary variables ≥ 1), so these too are set covering
constraints. Therefore, this crew scheduling problem is another example of a set covering
problem (where this particular set covering problem also includes the side constraint that
x1 + x2 + · · · + x12 ≤ 3).
Having identified the nature of the constraints, the stage now is set for formulating a BIP
spreadsheet model for this problem.
A BIP Spreadsheet Model for the Southwestern Airways Problem
Figure 7.6 shows a spreadsheet formulation of the complete BIP model for this problem. The
changing cells FlySequence? (C22:N22) contain the values of the 12 binary decision variables.
259
260 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
FIGURE 7.6
A spreadsheet formulation of the BIP model for the Southwestern Airways crew scheduling problem, where FlySequence (C22:N22)
shows the optimal solution obtained by Solver. The list of flight sequences under consideration is given in cells A25:D37.
A
1
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
Southwestern Airways Crew Scheduling Problem
2
Flight Sequence
3
4
1
2
3
4
5
6
7
8
9
10
11
12
5
2
3
4
6
7
5
7
8
9
9
8
9
Cost ($thousands)
At
6
Least
7
Includes Segment?
8
SFO–LAX
1
0
0
1
0
0
1
0
0
1
0
0
1
≥
1
Total
One
9
SFO–DEN
0
1
0
0
1
0
0
1
0
0
1
0
1
≥
1
10
SFO–SEA
0
0
1
0
0
1
0
0
1
0
0
1
1
≥
1
11
LAX–ORD
0
0
0
1
0
0
1
0
1
1
0
1
1
≥
1
12
LAX–SFO
1
0
0
0
0
1
0
0
0
1
1
0
1
≥
1
13
ORD–DEN
0
0
0
1
1
0
0
0
1
0
0
0
1
≥
1
14
ORD–SEA
0
0
0
0
0
0
1
1
0
1
1
1
1
≥
1
15
DEN–SFO
0
1
0
1
1
0
0
0
1
0
0
0
1
≥
1
16
DEN–ORD
0
0
0
0
1
0
0
1
0
0
1
0
1
≥
1
17
SEA–SFO
0
0
1
0
0
0
1
1
0
0
0
1
1
≥
1
18
SEA–LAX
0
0
0
0
0
1
0
0
1
1
1
1
1
≥
1
19
Total
20
21
22
Fly Sequence?
1
2
3
4
5
6
7
8
9
10
11
12 Sequences
0
0
1
1
0
0
0
0
0
0
1
0
3
Number
of Crews
≤
3
23
Total Cost ($thousands)
24
25
Flight Sequence Key
26
1
SFO-LAX
27
2
SFO-DEN-SFO
28
3
SFO-SEA-SFO
29
4
SFO-LAX-ORD-DEN-SFO
30
5
SFO-DEN-ORD-DEN-SFO
31
6
SFO-SEA-LAX-SFO
32
7
SFO-LAX-ORD-SEA-SFO
33
8
SFO-DEN-ORD-SEA-SFO
34
9
35
Solver Parameters
Set Objective Cell: TotalCost
To: Min
By Changing Variable Cells:
FlySequence?
Subject to the Constraints:
FlySequence? = binary
Total >= AtLeastOne
TotalSequences <= NumberOfCrews
18
O
7
Total
8
=SUMPRODUCT(C8:N8,FlySequence?)
9
=SUMPRODUCT(C9:N9,FlySequence?)
10
=SUMPRODUCT(C10:N10,FlySequence?)
11
=SUMPRODUCT(C11:N11,FlySequence?)
12
=SUMPRODUCT(C12:N12,FlySequence?)
13
=SUMPRODUCT(C13:N13,FlySequence?)
SFO-SEA-LAX-ORD-DEN-SFO
14
=SUMPRODUCT(C14:N14,FlySequence?)
10
SFO-LAX-ORD-SEA-LAX-SFO
15
=SUMPRODUCT(C15:N15,FlySequence?)
36
11
SFO-DEN-ORD-SEA-LAX-SFO
16
=SUMPRODUCT(C16:N16,FlySequence?)
37
12
SFO-SEA-LAX-ORD-SEA-SFO
17
=SUMPRODUCT(C17:N17,FlySequence?)
18
=SUMPRODUCT(C18:N18,FlySequence?)
Range Name
Cells
AtLeastOne
Cost
FlySequence?
IncludesSegment?
NumberOfCrews
Total
TotalCost
TotalSequences
Q8:Q18
C5:N5
C22:N22
C8:N18
Q22
O8:O18
Q24
O22
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
19
20
21
22
P
24
Total
Sequences
=SUM(FlySequence?)
Q
Total Cost ($thousands) =SUMPRODUCT(Cost,FlySequence?)
7.5 Using Mixed BIP to Deal with Setup Costs for Initiating Production: The Revised Wyndor Problem 261
The data in IncludesSegment? (C8:N18) and Cost (C5:N5) come directly from Table 7.5. The
last three columns of the spreadsheet are used to show the set covering constraints, Total ≥
AtLeastOne, and the side constraint, TotalSequences ≤ NumberOfCrews. Finally, the changing
cells FlySequence? have been constrained to be binary, as shown in the Solver Parameters box.
Solver provides the optimal solution shown in FlySequence? (C22:N22). In terms of the xj
variables, this solution is
x​ ​  3​​ = 1 (assign sequence 3 to a crew)
​  4​​ = 1​ ​  ​  (assign sequence 4 to a crew)​​ ​​
​​​x
​x​  11​​ = 1 (assign sequence 11 to a crew)
Many airlines are solving
huge BIP models of this
kind.
Review
Questions
and all other xj = 0, for a total cost of $18,000 as given by TotalCost (Q24). (Another optimal
solution is x1 = 1, x5 = 1, x12 = 1, and all other xj = 0.)
We should point out that this BIP model is a tiny one compared to the ones typically used
in actual practice. Airline crew scheduling problems involving thousands of possible flight
sequences now are being solved by using models similar to the one shown above but with
thousands of binary variables rather than just a dozen.
1. What is the crew scheduling problem that is encountered by companies in the travel industry?
2. What are the yes-or-no decisions that need to be made when addressing a crew scheduling
problem?
3. For the Southwestern Airways problem, there is a constraint for each flight to ensure that this
flight is covered by a crew. Describe the mathematical form of this constraint. Then explain in
words what this constraint is saying.
7.5 USING MIXED BIP TO DEAL WITH SETUP COSTS FOR INITIATING
PRODUCTION: THE REVISED WYNDOR PROBLEM
All of the examples considered thus far in this chapter have been pure BIP problems (problems where all the decision variables are binary variables). However, mixed BIP problems
(problems where only some of the decision variables are binary variables) also arise quite
frequently because only some of the decisions to be made are yes-or-no decisions and the rest
are how-much decisions.
One important example of this type is the product-mix problem introduced in Chapter 2,
but now with the added complication that a setup cost must be incurred to initiate the production of each product. Therefore, in addition to the how-much decisions of how much to
produce of each product, there also is a prior yes-or-no decision for each product of whether
to perform a setup to enable initiating its production.
To illustrate this type of problem, we will consider a revised version of the Wyndor Glass Co.
product-mix problem that was described in Section 2.1 and analyzed throughout most of Chapter 2.
The Revised Wyndor Problem with Setup Costs
Suppose now that the Wyndor Glass Co. will only devote one week each month to the production of the special doors and windows described in Section 2.1, so the question now is how
many doors and windows to produce during each of these week-long production runs. Since
the decisions to be made are no longer the rates of production for doors and windows, but
rather how many doors and windows to produce in individual production runs, these quantities now are required to be integer. (Note that this requirement means that these quantities
now will be represented by general integer variables that are not binary variables, but this will
not cause any complications when using Solver to solve the problem.)
Each time Wyndor’s plants convert from the production of other products to the production
of these doors and windows for a week, the following setup costs would be incurred to initiate
this production.
Setup cost to produce doors = $700
Setup cost to produce windows = $1,300
262 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions
Otherwise, all the original data given in Table 2.2 still apply, including a unit profit of $300
for doors and $500 for windows when disregarding these setup costs.
Table 7.6 shows the resulting net profit from producing any feasible quantity for either
product. Note that the large setup cost for either product makes it unprofitable to produce less
than three units of that product.
The dots in Figure 7.7 show the feasible solutions for this problem. By adding the appropriate entries in Table 7.6, the figure also shows the calculation of the total net profit P for
each of the corner points. The optimal solution turns out to be
​​(D, W) = (0, 6)​
with ​  P = 1,700​​
By contrast, the original solution
​​(D, W) = (2, 6)​
with ​  P = 1,600​​
now gives a smaller value of P. The reason that this original solution (which gave P = 3,600
for the original problem) is no longer optimal is that the setup costs reduce the total net profit
so much:
​P = 3,600 − 700 − 1,300 = 1,600​
FIGURE 7.7
The dots are the feasible
solutions for the revised
Wyndor problem. Also
shown is the calculation
of the total net profit P
(in dollars) for each corner
point from the net profits
given in Table 7.6.
Production W
quantity for
windows
8
(0, 6) gives P = 1,700
6
(2, 6) gives P = –100 + 1,700
= 1,600
4
(4, 3) gives P = 500 + 200
= 700
2
(0, 0)
gives P = 0
(4, 0) gives P = 500
0
2
4
Production quantity for doors
6
TABLE 7.6
Net Profit ($) for
the Revised Wyndor
Problem
8
D
Net Profit ($)
Number of
Units Produced
Doors
Windows
0
1
2
3
4
5
6
0 (300) − 0 = 0
1 (300) − 700 = −400
2 (300) − 700 = −100
3 (300) − 700 = 200
4 (300) − 700 = 500
Not feasible
Not feasible
0 (500) − 0 = 0
1 (500) − 1,300 = −800
2 (500) − 1,300 = −300
3 (500) − 1,300 = 200
4 (500) − 1,300 = 700
5 (500) − 1,300 = 1,200
6 (500) − 1,300 = 1,700
7.5 Using Mixed BIP to Deal with Setup Costs for Initiating Production: The Revised Wyndor Problem 263
Therefore, the graphical method for linear programming can no longer be used to find the
optimal solution for this new problem with setup costs.
How can we formulate a model for this problem so that it fits a standard kind of model that
can be solved by Solver? Table 7.6 shows that the net profit for either product is no longer
directly proportional to the number of units produced. Therefore, as it stands, the problem no
longer fits either linear programming or BIP. Before, for the original problem without setup
costs, the objective function was simply P = 300D + 500W. Now we need to subtract from
this expression each setup cost if the corresponding product will be produced, but we should
not subtract the setup cost if the product will not be produced. This is where binary variables
come to the rescue.
Formulation with Binary Variables
For each product, there is a yes-or-no decision regarding whether to perform the setup that
would enable initiating the production of the product, so the setup cost is incurred only if the
decision is yes. Therefore, we can introduce a binary variable for each setup cost and associate each value of the binary variable with one of the two possibilities for the setup cost. In
particular, let
These binary variables
enable subtracting each
setup cost only if the setup
is performed.
1, if perform the setup to produce doors
y​ ​  1​​ = ​​
​​​ ​
​​​
{0,
if not
​​ ​  ​
​ ​​
1, if perform the setup to produce windows
​y​  2​​ = ​​
​​​ ​
​​​
{0, if not
Therefore, the objective function now can be written as
​P = 300D + 500W − 700​y​  1​​ − 1,300​y​  2​​​
which fits the format for mixed BIP.
Since a setup is required to produce the corresponding product, these binary variables can
be related directly to the production quantities as follows:
1, if D > 0 can hold (can produce doors)
y​ ​  1​​ = ​​
​​​ ​
​
​​
{0, if D =